A. (1,2,6)
B. (2,1,6)
C. (6,2,1)
D. (6,1,2)
C2+C4+C6+……….. =
In the expression of (x4- (1/x3))15 coefficient of x32 is equal to :
If the first three terms of (1+ax)n are 1,6x,6x2 then (a,n)=
The coefficient of xk in the expansion of E=1+(1+x)+(1+x)2+……..+(1+x)n is
C0+3.C1+5.C2+……..+(2n+1).Cn =
The coefficient of x7 in (1+2x+3x2+4x3+……..∞)-3 is
C1/C0+2. C2/C1+3.C3/C2+....n. Cn/Cn-1=
Middle term in the expansion of (2x-3/x)15is
If the coefficients of 2nd, 3rd, 4th terms in the expansion of (1+x)2n are in A.P. then
If a is small in comparision with x, then (x/x+a)1/2+( x/x-a)1/2