Sum of Coefficients of All Terms
Trending Questions
Q.
Let be amatrix with . Let denote the row of .
If a matrix is obtained by performing the operation on , then is equal to:
Q. The value of 2000C2+2000C5+2000C8+...+2000C2000=?
- 21999−13
- 21999+13
- 22000+13
- 22000−13
Q.
is equal to?
None of these
Q. If (1+x+x2)20=a0+a1x+a2x2+⋯+a40x40, then the value of a20−a21+a22−⋯−a219 is
- 12a20(1−a20)
- 12a20(1+a20)
- 12a220
- \N
Q. The value of 20C0+20C1+⋯+20C8 is
- 219−( 20C10+ 20C9)2
- 219−( 20C10+2× 20C9)2
- 219− 20C102
- 219−(2× 20C10+ 20C9)2
Q. If Ar, Br, Cr denotes the coefficients of xr in the expansion of (1+x)10, (x+1)20, (1+x)30 respectively, then the value of 10∑r=1Ar(B10Br−C10Ar) is
- B10−C10
- A10(B210−C10A10)
- 0
- C10−B10
Q. If 2n+1C0+2n+1C1+2n+1C2+⋯+2n+1Cn=411 then which of the following is/are correct ?
- nC0+nC12+nC222+⋯nCn2n=(32)11
- nC0+nC12+nC222+⋯nCn2n=(23)11
- nC0−nC12+nC222−⋯+(−1)nnCn2n=(12)11
- nC0−nC12+nC222−⋯+(−1)nnCn2n=(23)11
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. If the value of 12C2+ 13C3+ 14C4+...+ 999C989 is 1000Ca+b, then |a+b|=
(where (a+b) is a single digit integer)
(where (a+b) is a single digit integer)
Q. If the expansion of 1(1−ax)(1−bx)=a0+a1x+a2x2+⋯+anxn+⋯, then an is (where a≠b, |ax|, |bx|<1)
- bn−anb−a
- an−bnb−a
- an+1−bn+1b−a
- bn+1−an+1b−a
Q. Let m, n∈N and gcd(2, n)=1. If 30(300)+29(301)+⋯+2(3028)+1(3029)=n⋅2m, then n+m is equal to
(Here(nk)=nCk)
(Here(nk)=nCk)
Q. If (1+x)n=C0+C1x+C2x2+…+Cnxn,
then the value of ∑∑0≤r<s≤n(Cr+Cs)2 is
then the value of ∑∑0≤r<s≤n(Cr+Cs)2 is
- (n−1)⋅2nCn−22n
- (n+1)⋅2nCn+22n
- (n+1)⋅2nCn−22n
- (n−1)⋅2nCn+22n
Q. The sum of last three digits of 7100−3100 is
Q. If (1+x)n=C0+C1x+⋯+Cnxn, then the value of n∑r=0n∑s=0CrCs is equal to
- 2n
- 22n
- 2n+1
- 23n
Q. If (x2006+1x2006+2)2010=a0+a1x+a2x2+....anxn, then the value of 2a0−a1−a2+2a3−a4−a5+2a6−... is
- 2
- 1
- 0
- 3
Q. If k and n be the two positive integers such that St=1t+2t+⋯+nt, then m∑r=1(m+1CrSr) is equal to:
- (n+1)m+1−(n+1)
- (n+1)m+1+(n+1)
- (n−1)m+1−(n−1)
- None of these
Q.
What is radians in degree?
Q. The value of 10C1+ 10C3+ 10C5+ 10C7+ 10C9 is
- 27
- 28
- 29
- 210
Q. 300∑r=0arxr=(1+x+x2+x3)100. If a=300∑r=0ar, then 300∑r=0rar is equal to
- 300a
- 100a
- 150a
- 75a
Q. If (1+x)n=C0+C1x+C2x2+…+Cnxn, then
the value of ∑∑0≤r<s≤n(r+s)CrCs is
the value of ∑∑0≤r<s≤n(r+s)CrCs is
- n[22n−1− 2n−1Cn−1]
- n[22n−1+ 2n−1Cn−1]
- 2n[22n−1− 2n−1Cn−1]
- 2n[22n−1+ 2n−1Cn−1]
Q. Sum of the binomial coefficients in the expansion of (a+b+c+d+e)n is
- n5
- n+1
- 5n
- 2n
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. If (1+x)n=C0+C1x+C2x2+⋯+Cnxn, then the value of ∑∑0≤r<s≤n(r⋅s)CrCs is
- n2[22n−3+122n−2Cn−1]
- [22n−3−122n−2Cn−1]
- n2[22n−3−122n−2Cn−1]
- [22n−3+122n−2Cn−1]
Q. Consider (1+x)2n+(1+2x+x2)n=2n∑r=0arxr, n∈N. If 2n∑r=0ar=f(n), then
- ∞∑n=11f(n)=16
- ∞∑n=11f(n)=38
- largest value of p for which f(5) is divisible by 2p is 11
- largest value of p for which f(5) is divisible by 2p is 9
Q. If (1+x)n=C0+C1x+…..+Cnxn, then the value of ∑∑0≤r<s≤nCrCs is equal to
- 12[22n−2nCn]
- 14[22n−2nCn]
- 12[22n+2nCn]
- 12[2n−2nCn]
Q. The sum 1⋅ 20C1−2⋅ 20C2+3⋅ 20C3−⋯−20⋅ 20C20 is equal to
Q. In the expansion of (1−x−x2+x3)6, the sum of the coefficients of x is
- 0
- 1
- 2
- 3
Q. Let (1+x)(1+x+x2)(1+x+x2+x3)......(1+x+x2+....+x30)=a0+a1x+a2x2+........+a465x465
then sum of a0+a2+a4+........+a464 is
then sum of a0+a2+a4+........+a464 is
- (31)!
- (31)!2
- (30)!
- (60)!2
Q. If a=31/223+1 and f(n)=n∑r=1(−1)r−1 nCr−1⋅an−r for all n≥3, then the value of f(2007)+f(2008) is
- 36
- 39
- 312
- 315
Q. If 2n+1C0+2n+1C1+2n+1C2+⋯+2n+1Cn=411 then which of the following is/are correct ?
- nC0+nC12+nC222+⋯nCn2n=(32)11
- nC0+nC12+nC222+⋯nCn2n=(23)11
- nC0−nC12+nC222−⋯+(−1)nnCn2n=(12)11
- nC0−nC12+nC222−⋯+(−1)nnCn2n=(23)11