If the first three terms of (1+ax)ⁿ are 1,6x,6x² then (a,n)=

The coefficient of x^k in the expansion of E=1+(1+x)+(1+x)²+……..+(1+x)ⁿ is

C₀+3.C₁+5.C₂+……..+(2n+1).C_n =

The coefficient of x⁷ in (1+2x+3x²+4x³+……..∞)^-3 is

C₁/C₀+2. C₂/C₁+3.C₃/C₂+....n. C_n/C_n-1=

Middle term in the expansion of (2x-3/x)¹⁵is

If the coefficients of 2^nd, 3^rd, 4^th 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

(√2+1)⁶+(√2-1)⁶=

3.C₀-5.C₁+7.C₂-9.C₃+…………(n+1) terms =