The sum of last eight consecutive coefficients in the expansion of $(1+x)^{15}$ is
$2^{15}$
$2^{14}$
$2^{16}$
$2^8$
$^n{C_0} - \frac{1}{2}{\,^n}{C_1} + \frac{1}{3}{\,^n}{C_2} - ...... + {( - 1)^n}\frac{{^n{C_n}}}{{n + 1}} = $
Let $\left(1+x+2 x^{2}\right)^{20}=a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{40} x^{40}$ then $a _{1}+ a _{3}+ a _{5}+\ldots+ a _{37}$ is equal to
The value of $-{ }^{15} C _{1}+2 .{ }^{15} C _{2}-3 .{ }^{15} C _{3}+\ldots \ldots$ $-15 .{ }^{15} C _{15}+{ }^{14} C _{1}+{ }^{14} C _{3}+{ }^{14} C _{5}+\ldots .+{ }^{14} C _{11}$ is
${C_0} - {C_1} + {C_2} - {C_3} + ..... + {( - 1)^n}{C_n}$ is equal to
If ${\sum\limits_{i = 1}^{20} {\left( {\frac{{{}^{20}{C_{i - 1}}}}{{{}^{20}{C_i} + {}^{20}{C_{i - 1}}}}} \right)} ^3}\, = \frac{k}{{21}}$, then $k$ equals