If (1-x3 )10= r=010arxr(1-x)30-2r, then 9a9a10 is equal to .
Binomial Theorem
Previous year questions with solutions · 2024 – 2026
Year-wise statistics questions · papers · average per paper · weightage
| Year | Questions | Papers | Avg / paper | Weightage |
|---|---|---|---|---|
| 2026 | 18 | 19 | 0.95 | 3.79% ▼ 21.7% vs last year |
| 2025 | 23 | 19 | 1.2 | 4.84% ▲ 38.3% vs last year |
| 2024 | 21 | 20 | 1.1 | 3.50% |
If the sum of the coefficients of x7 and x14 in the expansion of ( 1x3-x4 )n, x 0, is zero, then the value of n is .
If ( 115C0 + 115C1 ) ( 115C1 + 115C2 ) ( 115C12 + 115C13 ) = 1314C0 14C1 14C12, then 30 is equal to .
The sum of the series 2 1 20C4 - 3 2 20C5 + 4 3 20C6 - 5 4 20C7 + + 18 17 20C20, is equal to .
Let (1 + x + x2 )10 = a0 + a1x + a2x2 + + a20x20. If (a1 + a3 + a5 + + a19) - 11a2 = 121k, then k is equal to .
If = 1 + r=16(-3)r-1 12C2r-1, then the distance of the point (12, 3) from the line x - 3y + 1 = 0 is .
The sum of all rational terms in the expansion of (1+21/3+31/2 )6 is equal to .
If r=05 11C2r+12r+2= mn, (m,n)=1, then m-n is equal to .
If r=130 r2 (30Cr )230Cr-1= 229, then is equal to .
Let = r=0n (4r2 + 2r + 1 )nCr and = ( r=0n nCrr + 1 ) + 1n + 1. If 140 < 2 < 281, then the value of n is .
If the second, third and fourth terms in the expansion of (x + y)n are 135, 30 and 103, respectively, then 6 (n3 + x2 + y ) is equal to .
If the constant term in the expansion of (1 + 2x - 3x3 ) ( 32x2 - 13x )9 is p, then 108p is equal to .
If the coefficient of x30 in the expansion of (1+ 1x )6 (1+x2 )7 (1-x3 )8; x 0 is , then | | equals .
In the expansion of (1+x) (1-x2 ) (1+ 3x+ 3x2+ 1x3 )5, x 0, the sum of the coefficients of x3 and x-13 is equal to .
Let the coefficient of xr in the expansion of (x+3)n-1+(x+3)n-2(x+2)+(x+3)n-3(x+2)2+ +(x+2)n-1 be r. If r=0n r= n- n, , N, then the value of 2+ 2 equals .
Number of integral terms in the expansion of \7 ( 12 )+11 ( 16 ) \824 is equal to .
Let = k=0n ( ( n Ck )2k+1 ) and = k=0n-1 ( n Ck n Ck+1k+2 ). If 5 =6 , then n equals .
If 11C12+ 11C23+ + 11C910= nm with (n,m)=1, then n+m is equal to .
Remainder when 643232 is divided by 9 is equal to .
The coefficient of x2012 in the expansion of (1-x)2008 (1+x+x2 )2007 is equal to .