Sum of Coefficients of All Terms
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Q. The value of C0+2C1+3C2+4C3+…+(n+1)Cn is
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> ( where Cr= nCr)
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> ( where Cr= nCr)
- (n+2)⋅2n
- n⋅2n
- n⋅2n−1
- (n+2)⋅2n−1
Q. If (1+x+x2)20=a0+a1x+a2x2+⋯+a40x40, then the value of a0+3a1+5a2+⋯+81a40 is:
- 161×320
- 41×340
- 41×320
- 161×340
Q.
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Number of greatest binomial coefficients in the expansion of (1+x)2n and (1+x)2n+1 are p and q respectively (where n is a positive integer). Find the value of p+2q
Q. If Cr denotes coefficient of xr in (1+x)99, then the value of C0−2C1+3C2−4C3+⋯−100C99 is
Q.
11C01 + 11C12 + 11C23+.............. 11C1011 =
211−111
211−16
311−16
311−16
Q. The value of 15C0+ 15C1+ 15C2+…+ 15C7 is
- 214
- 215
- 213
- 27
Q.
Find the term independent of x in the expansion of (x2+2x4)(x+1x)32
Q. If the sum of the coefficients in the expansion of (a2x2−2ax+1)51 vanishes, then the value of a is
- 2
- −1
- 1
- −2
Q. The value of (21C1−10C1)+(21C2−10C2)+(21C3−10C3)+(21C4−10C4)+...+(21C10−10C10) is
- 221−211
- 221−210
- 220−29
- 220−210
Q. If n∑r=0(nCr−1nCr+ nCr−1)3=2524, then the value of n is
- 10
- 5
- 20
- 15
Q. Consider the expansion of (1+x)2n+1
If the coefficients of xr and xr+1 are equal in the expansion, then r is equal to
If the coefficients of xr and xr+1 are equal in the expansion, then r is equal to
- n
- 2n−12
- 2n+12
- n+1
Q. Any complex number in the polar form can be expressed in Euler's form as cosθ+isinθ=eiθ. This form of the complex number is useful in finding the sum of series n∑r=0 nCr(cosθ+isinθ)r.
n∑r=0 nCr(cosrθ+isinrθ)=n∑r=0 nCreirθ =n∑r=0 nCr(eiθ)r =(1+eiθ)n
Also, we know that the sum of binomial series does not change if r is replaced by n−r. Using these facts, answer the following questions.
If f(x)=50∑r=0 50Crsin2rx50∑r=0 50Crcos2rx, then the value of f(π8) is equal to
n∑r=0 nCr(cosrθ+isinrθ)=n∑r=0 nCreirθ =n∑r=0 nCr(eiθ)r =(1+eiθ)n
Also, we know that the sum of binomial series does not change if r is replaced by n−r. Using these facts, answer the following questions.
If f(x)=50∑r=0 50Crsin2rx50∑r=0 50Crcos2rx, then the value of f(π8) is equal to
- 1
- −1
- irrational value
- 0
Q. The value of 10C1+ 10C3+ 10C5+ 10C7+ 10C9 is
- 27
- 28
- 29
- 210
Q. If (1+x)n=C0+C1x+C2x2+…+Cnxn, then ∑∑0≤r<s≤n(r+s)(Cr+Cs+CrCs) is equal to
- n2⋅2n−n2[22n− 2nCn]
- n2⋅2n+n2[22n− 2nCn]
- n2⋅2n+n2[22n+ 2nCn]
- None of these
Q. The coefficient of x50 in (1+x)1000+x(1+x)999+x2(1+x)998+...+x1000 is
- 1001C50
- 1000C50
- 1001C51
- 1000C51
Q. The value of 19∑r=120Cr+1⋅(−1)r22r+1 is
- 2((34)20+4)
- −2((34)20+4)
- 2((34)20−4)
- −2((34)20−4)
Q. In the expansion of (512+718)1024, the number of integral terms is .
- 128
- 129
- 130
- 131
Q. Any complex number in the polar form can be expressed in Euler's form as cosθ+isinθ=eiθ. This form of the complex number is useful in finding the sum of series n∑r=0 nCr(cosθ+isinθ)r.
n∑r=0 nCr(cosrθ+isinrθ)=n∑r=0 nCreirθ =n∑r=0 nCr(eiθ)r =(1+eiθ)n
Also, we know that the sum of binomial series does not change if r is replaced by n−r. Using these facts, answer the following questions.
The value of 100∑r=0 100Cr(sinrx) is equal to
n∑r=0 nCr(cosrθ+isinrθ)=n∑r=0 nCreirθ =n∑r=0 nCr(eiθ)r =(1+eiθ)n
Also, we know that the sum of binomial series does not change if r is replaced by n−r. Using these facts, answer the following questions.
The value of 100∑r=0 100Cr(sinrx) is equal to
- 2100cos100(x2)sin50x
- 2100sin(50x)cos(x2)
- 2101cos100(50x)sin(x2)
- 2101sin100(50x)cos(50x)
Q. The coefficient of xn in the expansion of (1−x)(1−x)n is :
- n−1
- (−1)n(1+n)
- (−1)n−1(n−1)2
- (−1)n−1 n
Q. In the following equation given below, find the value of ′a′ :
4x2+ax+9=(2x+3)2
- a=11
- a=19
- a=17
- a=12
Q. The expression 22+122−1+32+132−1+42+142−1+..........+(2011)2+1(2011)2−1 lies in the interval
Q. For r=0, ⋯, 10, IfAr, Br and Cr denote respectively the coefficient of xr in the expansions of (1+x)10, (1+x)20, and (1+x)30. Then the value of ∑10r=1 Ar(B10Br−C10Ar) is
- B10−C10
- A10(B210−c10A10)
- \N
- C10−B10
Q. n∑r=0r2.(nCr)2 is equal to
- n2 . 2n−2Cn−1
- n2 . 2nCn
- n2 . 2n−1(r=1∏n−1 (2r−1))(n−1)!
- 2n (n∏r=1(2r−1))n!
Q. The value of n∑r=0(−2)r(nCrr+2Cr) is
- 1n+1, when n is odd
- 1n+2, when n is odd
- 1n+1, when n is even
- 1n+2, when n is even