Eamcet Mathematics - Binomial Theorem & Its Application Chapter Sample Question Paper With Answer Key
Exam Duration: 60 Mins Total Questions : 50
The sum of the coefficients in the expansion of\((1+2x-x^2)^{2143}\)is
- (a)
1
- (b)
\(2^{2143}\)
- (c)
\(3^{2143}\)
- (d)
0
To find sum of coeffiecnts put x=1
Required sum =(1+2-1)2143=22143
In the expansion of (1+x)50,the sum of the coefficients of odd power of x is
- (a)
0
- (b)
249
- (c)
250
- (d)
251
In the expansion of (1+x)n:
Sum of odd coefficients =sum of even coefficients
=2n-1
Here n=5
Therefore required sum=249
The number of the expansion of (a+b+c)n is
- (a)
n+1
- (b)
n+3
- (c)
\((n+1)(n+2)\over2\)
- (d)
None of these
Expanding [a+(b+c)]n, we get
[a+(b+c)]n=an+nC1.an-1(b+c) + nC2.an-2(b+c)2 + nC2.an-3(b+c)3+ ... +nCn.(b+c)n
Obviously the first term in the expansion gives 1 term, second term gives 2 terms, third term gives 3 terms and the last term gives (n + 1) terms.
∴ Total number of terms 1+2+3 + .... + (n +1)
\(={(n+1)(n+2)\over 2}\)
Cofficient of x2 in the expansion of(1+4x+x2)1/2 is
- (a)
-3
- (b)
-1
- (c)
2
- (d)
None of these
Expansion of
\((1+4x+2x^2)^{1/2}=1+{1\over 2}(4x+2x^2)+{{1\over 2}\left({1\over 2}-1\right)\over 12}.(4x+2x^2)^2+...\)
∴ Coefficient of \(x^2=1-{1\over 8}\times16=-1\)
The greatest value of n for which 3n occurs in 100!,is
- (a)
33
- (b)
44
- (c)
48
- (d)
52
\(n=E_p(3)=\left[100\over 3\right]+\left[100\over 3^2\right]+\left[100\over 3^3\right]+\left[100\over 3^4\right]+\left[100\over 3^5\right]+.......\)
(where [ ] denotes•the greatest integer function)
Ep (3)=33+11+3+1=48
Hence, the highest value of x is 48.
Let \(R=\left( 5\sqrt { 5 } +11 \right) _{ }^{ 2n-1 }\)and f=R-[R],where [ ] denotes the greatest integer function,then Rf is equal to
- (a)
42n+1
- (b)
42n
- (c)
42n-1
- (d)
None of these
The coefficient of t29 in the expansion of(1+t2)12(1+t12)(1+t21) is
- (a)
\(_{ }^{ 12 }{ C }_{ 4 }^{ }{ }\)
- (b)
\(_{ }^{ 12 }{ C }_{ 5}^{ }{ }\)
- (c)
\(_{ }^{ 12 }{ C }_{ 6 }^{ }{ }\)
- (d)
\(_{ }^{ 12 }{ C }_{ 7}^{ }{ }\)
The value of \(\sum _{ r=0 }^{ n-1 }{ \frac { { { { { ^{ n }C } } } }_{ r } }{ { { { { ^{ n }C } } } }_{ r }+{ { { { ^{ n }C } } } }_{ r+1 } } } \) is
- (a)
\(\frac { n }{ 2 } \)
- (b)
\(\frac { n+1 }{ 2 } \)
- (c)
\(\frac { n-1 }{ 2 } \)
- (d)
\(2n\)
\(\sum_{r=0}^{n-1}{^nC_r\over ^nC_r+^nC_{r+1}}=\sum_{r=0}^{n-1}{^nC_r\over ^{n+1}C_r+^nC_{r+1}}=\sum_{r=0}^{n-1}{r+1\over n+1}\)
\(={2\over n+1}\sum_{r=0}^{n-1}(r+1)={1\over (n+1)}[12+3+...+n]\)
\(={n(n+1)\over 2(n+1)}={n\over2}\)
If the middle term of \(\left( \frac { 1 }{ x } +\ xsin\quad x \right) ^{ 10 }\) is equal to \({ 7 }\frac { 7 }{ 8 } \), then value of x is
- (a)
\({ 2n\pi + }\frac { \pi }{ 6 } \)
- (b)
\({ n\pi + }\frac { \pi }{ 6 } \)
- (c)
\({ n\pi +(-1)^{ n } }\frac { \pi }{ 6 } \)
- (d)
\({ n\pi +(-1)^{ n } }\frac { \pi }{ 3 } \)
Middle term = \(\left( \frac { 10 }{ 2 } +1 \right) \) th term = 6th term
\(\therefore \) \({ T }_{ 6 }=^{ 10 }{ C }_{ 5 }\frac { 1 }{ { x }^{ 5 } } .{ x }^{ 5 }.{ sin }^{ 5 }x=\frac { 63 }{ 8 } \Rightarrow { sin }^{ 5 }x=\left( \frac { 1 }{ 2 } \right) ^{ 5 }\)
\(\Rightarrow \quad sinx=\frac { 1 }{ 2 } \Rightarrow \quad x=n\pi +(-1)^{ n }\frac { \pi }{ 6 } \)
If numerically the greatest coefficient in the expansion of (1-x)n is the coefficient of the 5th term, then total number of terms are
- (a)
10
- (b)
9
- (c)
12
- (d)
11
Since, the coefficient of the middle term in the expansion of (1-x)n or (1+x)n is the numerically greatest coefficient greatest coefficient and its value is \(^{ n }{ C }_{ \frac { n }{ 2 } }or\left[ ^{ n }{ C }_{ \frac { n-1 }{ 2 } }=^{ n }{ C }_{ \frac { n+1 }{ 2 } } \right] \).Also, it is given that the coefficient of the 5th term is the greatest, so 5th term is middle term.
\(\therefore \) n=8 So, total number of terms are (n+1) = 8+1 = 9
The greatest term in the expansion of (3+5x)15, when x=\(\frac { 1 }{ 5 } \), is
- (a)
15C3
- (b)
15C4.311
- (c)
15C10.310
- (d)
15C3.313
The greatest term in (3+5x)15 is \(\frac { \left( 15+1 \right) \left| 5x \right| }{ \left| 3 \right| +\left| 5x \right| } =\frac { \left( 15+1 \right) 5\left( \frac { 1 }{ 5 } \right) }{ 3+5.\frac { 1 }{ 5 } } =\frac { 16\times 1 }{ 4 } =4\)
So, 4th and 5th terms are the greatest term.
\(\therefore \) T5 = 15C4.311
The value of \(^{ 50 }{ C }_{ 4 }+\overset { 6 }{ \underset { r=1 }{ \Sigma } } ^{ 56-r }{ C }_{ 3 }\) is
- (a)
56C4
- (b)
56C3
- (c)
55C3
- (d)
55C4
Now, 50C4 + 55C3 + 54C3 + 53C3 + 52C3 + 51C3 + 50C3
= 50C3 + 50C4 + 51C3 + 52C3 + 53C3 + 54C3 + 55C3
= 51C4 + 51C3 + 52C3 + 53C3 + 54C3 + 55C3 [ \(\because \) nCr + nCr-1 = n+1Cr]
= 52C4 + 52C3 + 53C3 + 54C3 + 55C3
= 53C4 + 53C3 + 54C3 + 55C3
= 54C4 + 54C3 + 55C3 = 55C4 + 55C3 = 56C4
If the coefficient of x in equals the coefficient of x in , then a and b satisfy the relation
- (a)
ab = 1
- (b)
a/b = 1
- (c)
a+b = 1
- (d)
a-b = 1
Let x7 be contained in (r+1)th term in the expansion of
\(\left(ax^2+{1\over bx}\right)^{11}\)
\(∴\ \ T_{r+1}=\ ^{11}C_r(ax^2)^{11-r}\left(1\over bx\right)^r=\ ^{11}C_r{a^{11-r}\over br}.x^{22-3r}\)
For coefficient of x7, put 22-3r=7 ⇒ r = 5
\(∴\ \ T_6=\ ^{11}C_5{a^6\over b^5}.x^7\)
∴ Coefficient of x7 in the expansion of \(\left(ax^2+{1\over bx}\right)^{11}\ is\ ^{11}C_5{a^6\over b^5 }\)
Similarly, coefficient of x-7 in the expansion of \(\left(ax-{1\over bx^2}\right)^{11}\ is\ ^{11}C_6{a^5\over b^6}\)
According to the given condition, \(^{11}C_5{a^6\over b^5}=\ ^{11}C_6{a^5\over b^6}\)
\(⇒\ {a^6\over b^5}={a^5\over b^6}\Rightarrow ab=1\)
If x is so small that x3 and higher powers of x may be neglected, then \(\frac { { \left( 1+x \right) }^{ 3/2 }-{ \left( 1+\frac { 1 }{ 2 } x \right) }^{ 3 } }{ { \left( 1-x \right) }^{ 1/2 } } \) may be approximated as
- (a)
\(\frac { x }{ 2 } -\frac { 3 }{ 8 } { x }^{ 2 }\)
- (b)
\(-\frac { 3 }{ 8 } { x }^{ 2 }\)
- (c)
\(3x+\frac { 3 }{ 8 } { x }^{ 2 }\)
- (d)
\(1-\frac { 3 }{ 8 } { x }^{ 2 }\)
\({(1+x)^{3/2}-\left(1+{1\over 2}x\right)\over (1-x)^{1/2}}\)
\(={\left(1+{3\over 2}x+{{3\over 2}.{1\over 2}\over 2}x^2\right)-\left(1+{3x\over2}+{3.2\over 2}.{x^2\over 4}\right)\over(1-x)^{1/2}}\)
\(=-{3x^2\over 8}(1-x)^{-1/2}=-{3x^2\over 8}\left(1+{3\over 2}x+{{3\over 2}.{1\over 2}\over 2}x^2\right)\)
\(=-{3x^2\over 8}\) [since, higher powers of x2 can be neglected]
The coefficient of the middle term in the binomial expansion in powers of x of (1+\(\alpha \)x) and of (1-\(\alpha \)x)6 is the same, if is equal to
- (a)
\(-\frac { 5 }{ 3 } \)
- (b)
\(\frac { 10 }{ 3 } \)
- (c)
\(-\frac { 3 }{ 10 } \)
- (d)
\(\frac { 3 }{ 5 } \)
The coefficient of the middle term in powers of x in the expansion of \(\left( 1+\alpha x \right) ^{ 4 }=^{ 4 }{ C }_{ 2 }{ \alpha }^{ 2 }\)
The coefficient of the middle term in powers of x in the expansion of \(\left( 1-\alpha x \right) ^{ 6 }=^{ 6 }{ C }_{ 3 }\left( -\alpha \right) ^{ 3 }\)
According to the given condition,
\(^{ 4 }{ C }_{ 2 }{ \alpha }^{ 2 }=^{ 6 }{ C }_{ 3 }\left( -\alpha \right) ^{ 3 }\quad \Rightarrow \frac { 4! }{ 2!2! } { \alpha }^{ 2 }=-\frac { 6! }{ 3!3! } { \alpha }^{ 3 }\)
\(\Rightarrow \quad 6{ \alpha }^{ 2 }=-20{ \alpha }^{ 3 }\quad \Rightarrow \quad { \alpha }=-\frac { 6 }{ 20 } \)
\(\therefore \quad \quad \quad \quad \alpha =-\frac { 3 }{ 10 } \)
If a, b, c are in AP, then the sum of the coefficients of {1 + (ax2 - 2bx + c)2}1973 is
- (a)
-2
- (b)
-1
- (c)
0
- (d)
1
\(\therefore\) a, b, c are in AP
\(\Rightarrow \) 2b = a+c
\(\Rightarrow \) a - 2b + c =0
Putting x = 1
Required sum = (1 + a -2b + c)1973 = (1 + 0)1973
= 1
If the three successive coefficients in the binomial expansion of{1 + x)n are 28, 56 and 70 respectively, then n equals
- (a)
4
- (b)
6
- (c)
8
- (d)
9
\(^{ n }{ { C }_{ r-1 } }=28,\quad ^{ n }{ { C }_{ r } }=56,\quad ^{ n }{ { C }_{ r+1 } }=70\)
\(\therefore \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \frac { ^{ n }{ { C }_{ r } } }{ ^{ n }{ { C }_{ r-1 } } } =2\)
\(\Rightarrow \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \frac { n-r+1 }{ r } =2\)
\(\Rightarrow \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad n-3r+1=0\)
and \(\frac { ^{ n }{ { C }_{ r+1 } } }{ ^{ n }{ { C }_{ r } } } =\frac { 70 }{ 56 } \)
\(\Rightarrow \quad \quad \quad \quad \quad \quad \quad \frac { n-\left( r+1 \right) +1 }{ r+1 } \quad =\frac { 5 }{ 4 } \)
\(\Rightarrow \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 4n-4r=5r+5\)
\(\Rightarrow \quad \quad \quad \quad \quad \quad \quad \quad 4n-9r-5=0\)
Solving Eqs. (i) and (ii), we get
n = 8, r = 3
The number 525 - 325 is divisible by
- (a)
2
- (b)
3
- (c)
5
- (d)
7
525 - 325
= (5 - 3) (524 + 523 .3 + 522. 32 +......+ 324)
\(\therefore\) 525 - 325 is divisible by 2
If the number of terms in the expansion of (1 + 2x - 3x2)n is 36, then n equals
- (a)
7
- (b)
8
- (c)
9
- (d)
none of these
n+2C2 = 36 \(\Rightarrow\) \(\frac { \left( n+2 \right) \left( n+1 \right) }{ 2 } =36\)
\(\Rightarrow\) (n + 2)(n + 1) = 72 = 9 x 8
\(\therefore\) n + 1 = 8 \(\Rightarrow\) n = 7
The expression (10C0)2 - (10C1)2 + .... - (10C9)2 + (10C10)2 equals
- (a)
10C5
- (b)
-10C5
- (c)
(10C5)2
- (d)
(10!)2
(1 + x)10 = 10C0 + 10C1x + 10C2x2 + .... + 10C9x9 + 10C10x10 ... (i)
and (x - 1)10 = 10C0x10 - 10C1x9 + 10C2x8 - ... - 10C9x + 10C10 ... (ii)
Multiplying Eq. (i) and Eq. (ii), then
(x2 - 1)10 = 10C0 + 10C1x + ..... + 10C9x9 + 10C10x10) x (10C0x10 - 10C1x9 + .... - 10C9x + 10C10)
Comparing the coefficient of x10, then
10C5(-1)5 = (10C0)2 - (10C1)2 + ... - (10C9)2 + (10C10)2
\(\therefore\) (10C0)2 - (10C1)2 + .... - (10C9)2 + (10C10)2 = -10C5
The coefficient of a3b4c in the expansion of (1 + a + b - c)9 is
- (a)
2. 9C7. 7C4
- (b)
-2. 9C2. 7C3
- (c)
9C7. 7C4
- (d)
none of these
Coefficient of a3b4c in (1 + a + b - C)9
\(=\frac { 9! }{ 3!4!1! } \left( -1 \right) ^{ 1 }\\ =-\frac { 9! }{ 2!7! } \times \frac { 7!2! }{ 3!4! } \)
= 9C2 x 7C3 x 2
= -2. 9C2. 7C3
The value of x, for which the 6th term in the expansion of \(\left\{ { 2 }^{ \log { \sqrt { \left( { 9 }^{ x-1 }+7 \right) } } }+\frac { 1 }{ { 2 }^{ \left( 1/5 \right) \log _{ 2 }{ \left( { 3 }^{ x-1 }+1 \right) } } } \right\} \) is 84, is equal to
- (a)
4
- (b)
3
- (c)
2
- (d)
5
Given, T6 = 84
\(\therefore\) T5+1 = 84
7C5\(\left( { 2 }^{ \log _{ 2 }{ \sqrt { { 9 }^{ x-1 }+7 } } } \right) ^{ 2 }\left( \frac { 1 }{ { 2 }^{ 1/5\log _{ 2 }{ \left( { 3 }^{ x-1 }+1 \right) } } } \right) ^{ 5 }=84\)
\(\Rightarrow \quad \quad 21\left\{ \sqrt { { 9 }^{ x-1 }+7 } \right\} ^{ 2 }\left( \frac { 1 }{ { 2 }^{ \log _{ 2 }{ \left( { 3 }^{ x-1 }+1 \right) } } } \right) =84\\ \Rightarrow \quad \quad \quad \left( { 9 }^{ x-1 }+7 \right) \left( \frac { 1 }{ { 3 }^{ x-1 }+1 } \right) =4\)
Put 3x-1 = t
\(\therefore\) \(\frac { { t }^{ 2 }+7 }{ t+1 } =4\)
\(\Rightarrow\) t2 - 4t +3 = 0
(t - 1)(t - 3) = 0
\(\therefore\) t = 1, 3
\(\Rightarrow\) 3x-1 = 1, 3x-1 = 3
\(\therefore\) x - 1 =0 and x - 1 = 1, \(\therefore\) x = 1, 2
The value of the sum of the series \(3.^{ n }{ C }_{ 0 }-8.^{ n }{ C }_{ 1 }+13.^{ n }{ C }_{ 2 }-18.^{ n }{ C }_{ 3 }+...\)upto (n+1) terms is
- (a)
0
- (b)
3n
- (c)
5n
- (d)
none of these
\(3.^{ n }{ C }_{ 0 }-8.^{ n }{ C }_{ 1 }+13.^{ n }{ C }_{ 2 }-18.^{ n }{ C }_{ 3 }+...\)
\(=\quad 3(^{ n }{ C }_{ 0 }-^{ n }{ C }_{ 1 }+^{ n }{ C }_{ 2 }-^{ n }{ C }_{ 3 }+...)+5(-^{ n }{ C }_{ 1 }+2.^{ n }{ C }_{ 2 }-3.^{ n }{ C }_{ 3 }+...)\)
\(\because \left( 1-x \right) ^{ n }=^{ n }{ C }_{ 0 }-^{ n }{ C }_{ 1 }x+^{ n }{ C }_{ 2 }{ x }^{ 2 }+^{ n }{ C }_{ 3 }{ x }^{ 3 }+....\qquad ...(i)\)
and \(-n\left( 1-x \right) ^{ n-1 }=-^{ n }{ C }_{ 1 }+2^{ n }{ C }_{ 2 }x-3^{ n }{ C }_{ 3 }{ x }^{ 2 }+...\quad \quad ...(ii)\)
Putting x = 1 in Eqs. (i) and (ii) we get
\(\quad ^{ n }{ C }_{ 0 }-^{ n }{ C }_{ 1 }+^{ n }{ C }_{ 2 }-^{ n }{ C }_{ 3 }+...=0\)
and \(-^{ n }{ C }_{ 1 }+2.^{ n }{ C }_{ 2 }-3.^{ n }{ C }_{ 3 }+...=0\)
Hence \(3.^{ n }{ C }_{ 0 }-8.^{ n }{ C }_{ 1 }+13.^{ n }{ C }_{ 2 }-18.^{ n }{ C }_{ 3 }+...=0\)
For \(1\le r\le n\) the value of nCr + n-1Cr + n-2Cr + ....+ rCr is
- (a)
nCr+1
- (b)
n+1Cr
- (c)
n+1Cr+1
- (d)
none of these
nCr + n-1Cr + n-2Cr + .... + rCr
= Coefficient of xr in
{(I + x)n + (1 + x)n-1 + (1 + x)n-2 + ... + (1 + x)r}
= Coefficient of xr in \(\left\{ \frac { { \left( 1+x \right) }^{ r }{ \left( 1+x \right) }^{ n-r+1 }-1 }{ \left( 1+x-1 \right) } \right\} \)
\(\Rightarrow\) Coefficient of xr+1 in {(1 + x)n+1 - (1 + x)r}
= n+1Cr+1
If \(\left( 6\sqrt { 6 } +14 \right) ^{ n+1 }=m\) and if f is the fractional part of m, then mf is equal to
- (a)
15n+1
- (b)
20n+1
- (c)
25n
- (d)
none of these
\(\left( 6\sqrt { 6 } +14 \right) ^{ n+1 }=m\) = [m] + f
\(0\le f<1\)
and let \({ f }^{ ' }=\left( 6\sqrt { 6 } +14 \right) ^{ n+1 }\)
0 < f' < 1
\(\therefore\) -1 < f - f' < 1
From Eq. (iii), f - f' = 0
f = f'
Then, mf = mf' = \(\left( 6\sqrt { 6 } -14 \right) ^{ n+1 }\left( 6\sqrt { 6 } +14 \right) ^{ n+1 }\)
= (216 - 196)n+1
= (20)n+1
If (1 + x + 2x2)20 = a0 + a1x + a2x2 + .... + a40x40, then a1 + a3 + a5 + .... + a37 equals
- (a)
219(220 - 21)
- (b)
220(219 - 19)
- (c)
219(220 + 21)
- (d)
none of these
\(\therefore\) (1 + x + 2x2)20 = a0 + a1x + a2x2 + a3x3 + a4x4 + a5x5+ .... + a37x37 + a38x38 + a39x39 + a40x40
Putting x = 1 and x = -1 and subtracting, we get
420 - 220 = 2 (a1 + a3 + a5 + ... + a37 + a39)
or a1 + a3 + a5 + ... + a37 + a39 = 219(220 - 1)
a1 + a3 + a5 + ... + a37 = 219(220 -1)-a39 ........ (i)
Now, \({ \left( 1+x+{ 2x }^{ 2 } \right) }^{ 20 }=\frac { 20! }{ \alpha !\beta !\gamma ! } { \left( 1 \right) }^{ \alpha }{ \left( x \right) }^{ \beta }{ \left( { 2x }^{ 2 } \right) }^{ \gamma }\)
For \({ \left( 1+x+{ 2x }^{ 2 } \right) }^{ 20 }=\frac { 20! }{ \alpha !\beta !\gamma ! } { \left( 1 \right) }^{ \alpha }{ \left( x \right) }^{ \beta }{ \left( { 2x }^{ 2 } \right) }^{ \gamma }\)
\(\quad \quad \quad \quad \quad \quad \gamma =19,\beta =1,\alpha =0\\ \therefore \quad \quad \quad \quad { a }_{ 39 }=\frac { 20! }{ 0!1!19! } .{ 2 }^{ 19 }=20.{ 2 }^{ 19 }\)
From Eq. (i) a1 + a3 + a5 + ...... + a37 = 219(220 - 21)
If (1+ x + X2)n == a0 + a1x + a2x2 + .... + a2nx2n, then
- (a)
a0 - a2 + a4 - a6 + ........= 0, if n is odd
- (b)
a1 - a3 + a5 - a7 + .........= 0, if n is even
- (c)
a0 - a2 + a4 - a6 +........=0, if n = 4p, \(p\varepsilon { I }^{ + }\)
- (d)
a1 - a3 + a5 - a7 +......=0, if 4p + 1, \(p\varepsilon { I }^{ + }\)
\(\therefore\) (1 + x + x2) = a0 + a1x + a2x2 + ... + a2n x2n
Puting x = i\(\left( i=\sqrt { -1 } \right) \)
Then, we get
(1 + i + i2)n = (a0 - a2 + a4 - a6 + ... ) + i (a1 - a3 + a5 - a7 + .... )
\(\Rightarrow\) in = (a0 - a2 + a4 - a6 + ... ) + i (a1 - a3 + a5 - a7 + .... )
If n is odd, then Re(in) = 0
\(\Rightarrow\) a0 - a2 + a4 - a6 + ... = 0
If n is even, then Im(in) = 0
\(\Rightarrow\) a1 - a3 + a5 - a7 + ... = 0
S1 = \(\overset { n }{ \underset { i=j }{ \Sigma } } \) a1 + a2 + a3 + .......... +an
S2 = \(\underset { 1\le i< }{ \Sigma } \underset { j\le n }{ \Sigma } \) ai aj = a1a2 + a1a3 + ........... +an-1an
S3 = \(\underset { 1\le i< }{ \Sigma } \underset { j\le k }{ \Sigma } \underset { \le n }{ \Sigma } \) aiajak = a1a2a3 + a1a2a4 + ................ + an-2an-1an
...........................................................
Sn = a1 a2 a3 ......an
Then, (x + a1)(x + a2)(x + a3).........................(x + an) can be written as xn + S1 xn-1 + S2 xn-2 +.......+Sn
If (1+x)n = C0 + C1x + C2x2 + C3x3 + ..............+Cnxn, then the coefficient of xn-1 in the expression
(x +C0)( x + C1) ( x + C2) ( x + C3) ...........(x + Cn) is
- (a)
\({ 2 }^{ 2n }-\frac { 2n! }{ { 2\left( n! \right) }^{ 2 } } \)
- (b)
\({ 2 }^{ 2n-1 }-\frac { 2n! }{ { 2\left( n! \right) }^{ 2 } } \)
- (c)
\({ 2 }^{ 2n+1 }-\frac { 2n! }{ { 2\left( n! \right) }^{ 2 } } \)
- (d)
\({ 2 }^{ 2n-2 }-\frac { 2n! }{ { 2\left( n! \right) }^{ 2 } } \)
\(\therefore\) (x + C0)(x + C1)(x + C2) .............. (x + Cn)
= \({ x }^{ n+1 }+\left( \overset { n }{ \underset { r=0 }{ \Sigma } } { C }_{ r } \right) { x }^{ n }+\left( \underset { 0\le i< }{ \Sigma } \underset { j\le n }{ \Sigma } { C }_{ i }{ C }_{ j } \right) { x }^{ n+1 }+......\)
Coefficient of xn-1 is \(\underset { 0\le i< }{ \Sigma } \underset { j\le n }{ \Sigma } { C }_{ i }{ C }_{ j }\)
= \(\frac { 1 }{ 2 } \left\{ \left( \overset { n }{ \underset { r=0 }{ \Sigma } } { C }_{ r } \right) ^{ 2 }-\left( \overset { n }{ \underset { r=0 }{ \Sigma } } { C }_{ r }^{ 2 } \right) \right\} \)
= \(\frac { 1 }{ 2 } \left\{ { 2 }^{ 2n }-^{ 2n }{ { C }_{ n } } \right\} \)
= \({ 2 }^{ 2n-1 }-\frac { 2n! }{ 2(n!)^{ 2 } } \)
S1 = \(\overset { n }{ \underset { i=j }{ \Sigma } } \) a1 + a2 + a3 + .......... +an
S2 = \(\underset { 1\le i< }{ \Sigma } \underset { j\le n }{ \Sigma } \) ai aj = a1a2 + a1a3 + ........... +an-1an
S3 = \(\underset { 1\le i< }{ \Sigma } \underset { j\le k }{ \Sigma } \underset { \le n }{ \Sigma } \) aiajak = a1a2a3 + a1a2a4 + ................ + an-2an-1an
...........................................................
Sn = a1 a2 a3 ......an
Then, (x + a1)(x + a2)(x + a3).........................(x + an) can be written as xn + S1 xn-1 + S2 xn-2 +.......+Sn
The coefficient of x6 (2 + x )3 (3 + x)2(5 + X)3 is in the expression
- (a)
78
- (b)
156
- (c)
312
- (d)
624
(x + 2)3 (x + 3)2 (x + 5)3 = (x + 2) (x + 2) (x + 2) (x + 3) (x + 3) x (x + 5) (x + 5) (x + 5)
= x8 + (2 + 2 + 2 + 3 + 3 + 5 + 5 + 5) x7 + x6(2· 2 + 2·2 + 2· 3 + ... + 5· 5)+ ....
\(\therefore\) Coefficient of x6 = \(\frac { { \left( 2+2+2+3+3+5+5+5 \right) }^{ 2 }-\left( { 2 }^{ 2 }+{ 2 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+{ 5 }^{ 2 }+{ 5 }^{ 2 } \right) }{ 2 } \)
= \(\frac { 729-105 }{ 2 } =312\)
\(\left( \sqrt { 3 } +1 \right) ^{ 4 }+\left( \sqrt { 3 } -1 \right) ^{ 4 }\) is a/an
- (a)
rational number
- (b)
irrational number
- (c)
negative number
- (d)
None of these
We have, \(\left( \sqrt { 3 } +1 \right) ^{ 4 }+\left( \sqrt { 3 } -1 \right) ^{ 4 }\)
= \(2\left( { 4 }_{ { C }_{ 0 } }\left( \sqrt { 3 } \right) ^{ 4 }+{ 4 }_{ { C }_{ 2 } }\left( \sqrt { 3 } \right) ^{ 2 }+{ 4 }_{ { C }_{ 4 } }\left( \sqrt { 3 } \right) ^{ 0 } \right) \)
= \(2(9+18+1)=56\)
If p be the sum of the of add terms and Q and of even terms in the expansion of (x + a)n, then the value of [(x + a)2n - (x -a)2n] equals
- (a)
PQ
- (b)
2PQ
- (c)
4 PQ
- (d)
None of these
(x + a)n = (t1 + t3 + t5 + ...) + (t2 + t4 + t6 + ....)
\(\therefore\) (x + a)n = P + Q ...............(i)
and (x - a)n = (t1 + t3 + t5 +.... ) - (t2 + t4 + t6 + ....)
(x - a)n = P - Q ..................(ii)
Squaring and subtracting (ii) from (i), we get,
(x +a)2n - (x - a)2n = (P + Q)2 - (P -Q)2 = 4PQ
if (1 + x + 2x2)6 = 1 + a1x + a2X2 + ... + a12X12, then the expression a2 + a4 + a6 + ... a12 has the value
- (a)
32
- (b)
63
- (c)
64
- (d)
31
(1 + x - 2x2)6 = 1 + a1x +a2X2 + .. + a12x12
On putting x = 1 and x = -1 in the given expansion and adding the results we get
64 = 2 (1 + a2 + a4 + ....a12)
\(\Rightarrow\) a2 + a4 + a6 + ... + a12 = 31
The digit at the units place in the number 192005 + 112005 - 92005 is
- (a)
2
- (b)
1
- (c)
0
- (d)
8
192005 + 112005 - 92005
Digits a units place in power of 19,11 and 9 is as follows:
191 = 9 111 = 1 91 = 9
192 = 1 112 = 1 92 = 9
193 = 9 113 =1 93 = 9
: : :
192005 = 9 112005 = 1 92005 = 9
So digit at units place 9 + 1 - 9 = 1
If n is a postive integer , then 52n + 2 - 24 n - 25 is divisible by
- (a)
574
- (b)
575
- (c)
675
- (d)
576
52n+2 - 24n - 25
= 52(52)n - 24n - 25 = 25 (25)n - 24n - 25
= 25 (24 + 1)n - 24n - 25
= 25 [1 + nC1 24 + nC2 242 + ...+nCn 24n] - 24n - 25
= 25 + 25 x24n + 25 [nC2 242 + .... + 24n] -24n - 25
24.24n + 25 [nC2 242 +...+24n]
= 576 [n + 25 {nC2 + ...24n-2}
\(\therefore\)52n+2 - 24n - 25 is divisble by 576
24n+4 - 15n -16, where n \(\epsilon\) N is divisible by
- (a)
225
- (b)
227
- (c)
229
- (d)
285
we have , 24n+4 - 15n - 16 = 24 (n +1) - 15n - 16
= 16n+1 - 15n - 16 = (1 + 15)n+1 - 15n - 16
=n+1C0150 + n+1c1151 + n+1C2152 + n+1C3153 +n+1Cn+1 (15)n+1 - 15n - 16
= 1 + 15n + 15 + n+1C2152 + n+1C3153 + n+1Cn+1 (15)n+1 - 15n - 16
= 152 [n+1C2 + n+1C315 + ...+(15)n-1]
Thus, 24n+4 - 15n - 16 is divisible by 225.
Find the term indepent of x in the expansion of \(\left( \frac { 3 }{ 2 } { x }^{ 2 }-\frac { 1 }{ 3x } \right) ^{ 6 }\)
- (a)
1/12
- (b)
7/12
- (c)
5/12
- (d)
None of these
Let (r + 1)th term is indepent of x
We have Tr+1 = 6Cr \(\left( \frac { 3 }{ 2 } { x }^{ 2 }-\frac { 1 }{ 3x } \right) ^{ 6 }\)
= 6Cr \(\left( \frac { 3 }{ 2 } \right) ^{ 6-r }({ x }^{ 2 })^{ 6-r }(-1)^{ r }\left( \frac { 1 }{ x } \right) r\left( \frac { 1 }{ { 3 }^{ r } } \right) \)
= (-1)r \(^{ \ 6 }{ C }_{ r },\frac { \left( 3 \right) ^{ 6-2r } }{ \left( 2 \right) ^{ 6-r } } { x }^{ 12-3r }\)
The term will be independent of x if the index of x is zero i.e., 12 - 3r = 0 thus r = 4
T4+1 = T5 = (-1)4 x \(^{ 6 }{ C }_{ r },\frac { \left( 3 \right) ^{ 6-8r } }{ \left( 2 \right) ^{ 6-4 } } =\frac { 5 }{ 12 } \)
Find the Coefficient of x11 in the expansion of \(\left( { x }^{ 3 }-\frac { 2 }{ { x }^{ 2 } } \right) ^{ 12 }\)
- (a)
-25344
- (b)
-25250
- (c)
-25000
- (d)
24310
Let (r + 1)th term contain x11
We have , Tr+1 = 12Cr (x3)12-r \(\left( \frac { 2 }{ { x }^{ 2 } } \right) ^{ r }\)
= 12Crx36-3r-2r (-1)r 2r
= 12 Cr (-1)r 2r x36-5r
Now for this to contain x11,36 - 5r = 11, i.e r = 5
Thus the coefficient of x11 is 12C5 (-1)525 = -25344
Find the middle terms in the expansion \(\left( \frac { P }{ x } +\frac { x }{ P } \right) ^{ 9 }\).
- (a)
\(\frac { 125p }{ x } ,\frac { 124x }{ P } \)
- (b)
\(\frac { 126p }{ x } ,\frac { 126x }{ p } \)
- (c)
\(\frac { 124p }{ x } ,\frac { 120x }{ p } \)
- (d)
\(\frac { 126p }{ x } ,\frac { 125x }{ p } \)
Since the power of the given expansion id odd therefore, we have two middle terms which are 5th and 6th terms.
\(\therefore\) \({ T }_{ 5 }=^{ 9 }{ C }_{ 4 }\left( \frac { P }{ x } \right) ^{ 5 }\left( \frac { x }{ p } \right) ^{ 4 }=^{ 9 }{ C }_{ 4 }\frac { p }{ x } =\frac { 126p }{ x } \)
\({ T }_{ 6 }=^{ 9 }{ C }_{ 5 }\left( \frac { P }{ x } \right) ^{ 5 }\left( \frac { x }{ p } \right) ^{ 5 }=^{ 9 }{ C }_{ 5 }\frac { x }{ p } =\frac { 126x }{ p } \)
the coefficient of x9 in th expansion of (1 + x + x2 + x3)3 (1-x)6 is
- (a)
-7
- (b)
7
- (c)
9
- (d)
-9
Given expansion can be written as,
(1 + x + x2 +x3)3 (1-x)3 (1-x)3
= (1-x4)3 (1-x)3
= (1-3x4+3x8 - x12) (1-3x +3x2 -x3)
\(\therefore\) The coefficient of x9 is 3(-3) = -9
If the sum of the coefficients in the expansion of (p + q)n is 1024, then the gretest coefficients in the expansion is,
- (a)
10C5
- (b)
10C4
- (c)
10C2
- (d)
10C9
Since sum of the coefficients in expansion of (p +q)n = 1024
\(\therefore\) 2n = 210 \(\Rightarrow\) n = 10
Which is even , so gretest coeffcient of (p + q)10 is coefficient of most middle term which is nCn/2 = 10C5
The middle term in in th expansion of \(\left( x+\frac { 1 }{ x } \right) ^{ 10 }\) is
- (a)
10C1\(\frac { 1 }{ x } \)
- (b)
10C5
- (c)
10C6
- (d)
10C7x
Middle term
= \({ T }_{ \frac { 10+2 }{ 2 } }={ T }_{ 5+1 }=^{ 10 }{ C }_{ 5 }{ x }^{ 10-5 }\left( \frac { 1 }{ 5 } \right) ^{ 5 }=^{ 10 }{ C }_{ 5 }\)
The coefficient of x7 in the expansion of \(\left( \frac { { x }^{ 2 } }{ 2 } -\frac { 2 }{ x } \right) ^{ 9 }\) is
- (a)
-56
- (b)
14
- (c)
-14
- (d)
None of these
\({ T }_{ r+1 }=^{ 9 }{ C }_{ r }\left( \frac { { x }^{ 2 } }{ 2 } \right) ^{ 9-r }\left( -\frac { 2 }{ x } \right) ^{ 5 }\)
= \(^{ 9 }{ C }_{ r }\left( \frac { 1 }{ 2 } \right) ^{ 9-r }(-2)^{ r }{ x }^{ 18-3r }\)
Now, for the coefficient of x7 , we put 18 - 3r = 7
\(\therefore\) r = 11/3 (not possible)
So, there is no term containing x7
The number of terms in the expansion of (a + b +c)10 is
- (a)
11
- (b)
21
- (c)
55
- (d)
66
We have (a + b + c)10 = [a + (b + c)]10
= 10C010 + 10C1a9(b + c) + 10C2a8 (b+ c)2 + ... + 10C10 (b + c)10
So, we can see that first term contains 1 term, seconds term contains 2 terms , 3rd term contaions 3 terms and so on
\(\therefore\) Total number of terms in the given expansion
= 1 + 2 +3 + ...+ 11 = \(\frac { 11(11+1) }{ 2 } =66\)
C0Cr +C1Cr+1 + C2Cr+2 + ...+Cn-rCn is equal to
- (a)
\(\frac { \left( 2n \right) ! }{ \left( n-r \right) !\left( n+r \right) ! } \)
- (b)
\(\frac { n! }{ r!\left( n+r \right) ! } \)
- (c)
\(\frac { n! }{ \left( n-r \right) ! } \)
- (d)
None of these
\(\because\) (1 + x)n = C0 + C1x + C2X2 + ..+CrXr + ..... CnXn ......(i)
and \(\left( 1+\frac { 1 }{ x } \right) ^{ n }={ C }_{ 0 }+{ C }_{ 1 }\frac { 1 }{ x } +{ C }_{ 2 }\frac { 1 }{ { x }^{ 2 } } +..+{ C }_{ r }\frac { 1 }{ { x }^{ r } } +{ C }_{ r+1 }\frac { 1 }{ { x }^{ r+1 } } +{ C }_{ r+2 }\frac { 1 }{ { x }^{ r+2 } } +..+{ C }_{ n }\frac { 1 }{ { X }^{ n } } ...(ii)\)
On multiplying (i) and (ii), equating coefficient of xr in \(\frac { 1 }{ { x }^{ n } } \left( 1+x \right) ^{ 2n }\) or the coefficient of xn+r (1 + x)2n , we get the value of required expression which is
\(^{ 2n }{ C }_{ n+r }=\frac { \left( 2n \right) ! }{ \left( n-r \right) !\left( n+r \right) ! } \)
If p is a real number and if the middle term in the expansion of \(\left( \frac { P }{ 2 } +2 \right) ^{ 8 }\) is 1120, then the values of p
- (a)
\(\pm\) 3
- (b)
\(\pm\) 1
- (c)
\(\pm\) 2
- (d)
None of these
Given expansion is \(\left( \frac { P }{ 2 } +2 \right) ^{ 8 }\) Here, n = 8 (even)
\(\Rightarrow\) Middle term \(\left( \frac { 8 }{ 2 } +1 \right) \)th = 5th
\(\therefore\) T5 = 8C4 \(\left( \frac { P }{ 2 } \right) ^{ 8-4 }\) . (2)4 = 1120
\(\Rightarrow\) P4 = 16 \(\therefore\) P = \(\pm\) 2
The 4th term in the expansion of (x - 2y)12 is
- (a)
1760x3y9
- (b)
-1760x9y3
- (c)
1760x9y3
- (d)
None of these
4th term in the expansion of (x - 2y)12 is
T4 =T3 + 1 = 12C3X12-3 (-2y)3
\({ T }_{ 4 }=\frac { 12\times 11\times 10 }{ 6 } { x }^{ 9 }\) x9 (-2)3 y3 = 2 x 11 x 10x9 (-8)y3
= 220 x (-8) x9y3 = -1760x9y3
Statement I : The term Independent of x in the expansion of \(\left( x+\frac { 1 }{ x } +2 \right) ^{ m }\) is \(\frac { \left( 4m \right) ! }{ (2m!)^{ 2 } } \)
Statement II : The coefficient of x6 in the expansion of (1 +x)n is nC6
- (a)
If both Statement-I and Statement-II are true and Staternent-Il is the correct explanation of Statement -I.
- (b)
If both Statement-I and Statement-Il are true but Statement-II is not the correct explanation of Statement -I.
- (c)
If Statement-I is true but Statement-II is false.
- (d)
If Statement -I is false and Statement-II is true.
\(\left( x+\frac { 1 }{ x } +2 \right) ^{ m } =\left( \frac { { x }^{ 2 }+2x+1 }{ x } \right) =\frac { \left( 1+x \right) ^{ 2m } }{ { x }^{ m } } \)
Term independent of x is coefficient of xm in the expansion of (1 + x)2m = 2mCm = \(\frac { \left( 2m \right) ! }{ (m!)^{ 2 } } \)
Coefficient of x6 in the expansion of (1 +x)n is nC6
Statement -I : if an = \(\sum _{ r=0 }^{ n }{ \frac { 1 }{ ^{ n }{ C }_{ r } } } \), then \(\sum _{ r=0 }^{ n }{ \frac { 1 }{ ^{ n }{ C }_{ r } } } \) is equal to \(\frac { n }{ 2 } \)an.
Statement II : \(^{ n }{ C }_{ r }=^{ n }{ C }_{ n-r }\)
- (a)
If both Statement-I and Statement-II are true and Staternent-Il is the correct explanation of Statement -I.
- (b)
If both Statement-I and Statement-Il are true but Statement-II is not the correct explanation of Statement -I.
- (c)
If Statement-I is true but Statement-II is false.
- (d)
If Statement -I is false and Statement-II is true.
Let b = \(\sum _{ r=0 }^{ n }{ \frac { 1 }{ ^{ n }{ C }_{ r } } } =\sum _{ r=0 }^{ n }{ \frac { n-(n-r) }{ ^{ n }{ C }_{ r } } } \)
= \(\sum _{ r=0 }^{ n }{ \frac { 1 }{ ^{ n }{ C }_{ r } } } -\sum _{ r=0 }^{ n }{ \frac { 1 }{ ^{ n }{ C }_{ r } } } \)
= \({ na }_{ n }-\sum _{ r=0 }^{ n }{ \frac { n-r }{ ^{ n }{ C }_{ n-r } } } \)
= nan - b \(\Rightarrow\) 2 b = nan \(\Rightarrow\) b = \(\frac { n }{ 2 } \)an
Statement I : The rth term from the end in the expansion of (x +a)n is nCn-r+1 xr-1 an-r+1
Statement II : The rth term from the end in the expansion of (x + a)n is (n - r + 2)th term from begining
- (a)
If both Statement-I and Statement-II are true and Staternent-Il is the correct explanation of Statement -I.
- (b)
If both Statement-I and Statement-Il are true but Statement-II is not the correct explanation of Statement -I.
- (c)
If Statement-I is true but Statement-II is false.
- (d)
If Statement -I is false and Statement-II is true.
rth term from the end in (x +a)n is same as (n - r + 2)th term from the begining which is given by
Tn-r + 2 = T(n - r + 1) + 1 = nCn -r + 1 xn(n - r + 1) an-r + 1
= nCn -r + 1 xr-1 an-r+1
Match the following.
Column - I | Column - II |
---|---|
(i) (1 - 2x)5 | (P) \(\frac { { x }^{ 5 } }{ 243 } -\frac { { 5x }^{ 3 } }{ 81 } +\frac { 10x }{ 27 } -\frac { 10 }{ 9x } +\frac { 5 }{ { 3x }^{ 3 } } -\frac { 1 }{ { x }^{ 5 } } \) |
(ii) (2x - 3)6 | (q) x6 + 6x4 + 15x2 + 20 + \(\frac { 15 }{ { x }^{ 2 } } +\frac { 6 }{ { x }^{ 4 } } +\frac { 1 }{ { x }^{ 6 } } \) |
(iii) \(\left( \frac { x }{ 3 } -\frac { 1 }{ x } \right) ^{ 5 }\) | 1 - 10x + 40x2 - 80x3 + 80x4 - 32x5 |
(iv) \(\left( x+\frac { 1 }{ x } \right) ^{ 6 }\) | 64x6 - 576x5 + 2160x4 - 4320x3 +4860x2 -2916x +729 |
- (a)
(i) \(\rightarrow\) (p), (ii) \(\rightarrow\) (q), (iii) \(\rightarrow\) (r), (iv)\(\rightarrow\) (s)
- (b)
(i) \(\rightarrow\)(r), (ii) \(\rightarrow\)(s), (iii) \(\rightarrow\) (p), (iv) \(\rightarrow\) (q)
- (c)
(i) \(\rightarrow\) (q), (ii) \(\rightarrow\) (s), (iii) \(\rightarrow\) (p), (iv) \(\rightarrow\) (r)
- (d)
(i) \(\rightarrow\) (r), (ii) \(\rightarrow\)(s), (iii) \(\rightarrow\)(q), (iv) \(\rightarrow\) (p)
(i) (1-2x)5 = 5C0(1)5 - 5C1(1)42x + 5C2(1)3 (2x)2 -5C3(1)2(2x)3 + 5C4(1)(2x)4-5C5(2x)5
=1 - 5 x 2x + \(\frac { 5\times 4 }{ 2 } \) x 4x2 - \(\frac { 5\times 4 }{ 2 } \) x 8x3 + 5 x 16x4 - 32x5
=1 -10x + 40x2 - 80x3 + 80x4 - 32x5
(ii) (2x -3)6 = 6C0(2x)6 - 6C1(2x)5 x3 + 6C2 (2x)4 (3)2 -6C3(2x)3 (3)3 +6C4(2x)2(3)4 - 6C5(2x)35 + 6C636
= 64x6 -6 x 32 x 3 x x5 + \(\frac { 6\times 5 }{ 2 } \) x16x4 x 9
= \(\frac { 6\times 5\times 4\times 8x^{ 3 }\times 27 }{ 6 } +\frac { 6\times 5 }{ 2 } \) x 4x2 x 81 - 6 x 2x 243 + 729
= 64x6 - 576x5 +2160x4 - 4320x3 + 4860x2 - 2916x + 729
(iii) \(\left( \frac { x }{ 3 } -\frac { 1 }{ x } \right) ^{ 5 }=^{ 5 }{ C }_{ 0 }\left( \frac { x }{ 3 } \right) ^{ 5 }-^{ 5 }{ C }_{ 1 }\left( \frac { x }{ 3 } \right) ^{ 4 }\frac { 1 }{ x } +^{ 5 }{ C }_{ 2 }\left( \frac { x }{ 3 } \right) ^{ 3 }\left( \frac { 1 }{ x } \right) ^{ 2 }-^{ 5 }{ C }_{ 3 }\left( \frac { x }{ 3 } \right) ^{ 2 }\left( \frac { 1 }{ x } \right) ^{ 3 }+^{ 5 }{ C }_{ 4 }\left( \frac { x }{ 3 } \right) \left( \frac { 1 }{ x } \right) ^{ 4 }-^{ 5 }{ C }_{ 5 }\left( \frac { 1 }{ 3 } \right) ^{ 5 }\)
= \(\frac { { x }^{ 5 } }{ 243 } -\frac { { 5x }^{ 3 } }{ 81 } +\frac { 10x }{ 27 } -\frac { 10 }{ 9x } +\frac { 5 }{ { 3x }^{ 3 } } -\frac { 1 }{ { x }^{ 5 } } \)
(iv) \(\left( x+\frac { 1 }{ x } \right) ^{ 6 }=^{ 6 }{ C }_{ 0 }{ x }^{ 6 }+^{ 6 }{ C }_{ 0 }{ x }^{ 5 }\left( \frac { 1 }{ x } \right) +^{ 6 }{ C }_{ 0 }{ x }^{ 4 }\left( \frac { 1 }{ x } \right) ^{ 2 }+^{ 6 }{ C }_{ 0 }{ x }^{ 3 }\left( \frac { 1 }{ x } \right) ^{ 3 }+^{ 6 }{ C }_{ 0 }{ x }^{ 2 }\left( \frac { 1 }{ x } \right) ^{ 4 }+^{ 6 }{ C }_{ 5 }{ x }\left( \frac { 1 }{ x } \right) ^{ 5 }+^{ 6 }{ C }_{ 6 }\left( \frac { 1 }{ x } \right) ^{ 6 }\)
= x6 + 6x4 + 15x2 + 20 + \(\frac { 15 }{ { x }^{ 2 } } +\frac { 6 }{ { x }^{ 4 } } + \frac { 1 }{ { x }^{ 6 } } \)