In the expansion of ${(1 + x)^5}$, the sum of the coefficient of the terms is

If the expansion in powers of $x$ of the function  $\frac{1}{{\left( {1 - ax} \right)\left( {1 - bx} \right)}}$ is ${a_0} + {a_1}x + {a_2}{x^2} + \;{a_3}{x^3} + \; \ldots......$ then  ${a_n}$ is

Co-efficient of $\alpha ^t$ in the expansion of,

$(\alpha + p)^{m - 1} + (\alpha + p)^{m - 2} (\alpha + q) + (\alpha + p)^{m - 3} (\alpha + q)^2 + ...... (\alpha + q)^{m - 1}$

where $\alpha \ne - q$ and $p \ne q$ is :

Let $(1+2 x)^{20}=a_0+a_1 x+a_2 x^2+\ldots+a_{20} x^{20}$.Then $3 a_0+2 a_1+3 a_2+2 a_3+3 a_4+2 a_5+\ldots+2 a_{19}+3 a_{20}$ equals

The sum of the series $\sum\limits_{r = 0}^n {{{( - 1)}^r}\,{\,^n}{C_r}\left( {\frac{1}{{{2^r}}} + \frac{{{3^r}}}{{{2^{2r}}}} + \frac{{{7^r}}}{{{2^{3r}}}} + \frac{{{{15}^r}}}{{{2^{4r}}}} + .....m\,{\rm{terms}}} \right)} $ is

The sum of the last eight coefficients in the expansion of ${(1 + x)^{15}}$ is