A. 2n
B. 3n
C. 4n
D. 5n
3.C0+7.C1+11.C2+……..+(4n+3).Cn =
(1.02)6 + (0.98)6 =
(2-)5 + (2+)5 =
If the 3rd, 4th and 5th terms of (x+a)n are 60, 160 and 240 respectively then (x,a,n)=
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