${(1 + x - 3{x^2})^{3148}}$ के विस्तार में गुणांकों का योगफल होगा

$x \in R , x \neq-1$ के लिए, यदि $(1+x)^{2016}+x(1+x)^{2015}+x^{2}(1+x)^{2014}$ $+\ldots .+x^{2016}=\sum_{i=0}^{2016} a_{i} x^{i}$ है, तो $a_{17}$ बराबर है

$\left( {\left( {\begin{array}{*{20}{c}}
{21}\\
1
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
1
\end{array}} \right)} \right) + \left( {\left( {\begin{array}{*{20}{c}}
{21}\\
2
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
2
\end{array}} \right)} \right)$$ + \left( {\left( {\begin{array}{*{20}{c}}
{21}\\
3
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
3
\end{array}} \right)} \right) + \;.\;.\;.$$ + \left( {\left( {\begin{array}{*{20}{c}}
{21}\\
{10}
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
{10}
\end{array}} \right)} \right)$ का मान है:

माना $m, n \in N$ तथा $\operatorname{gcd}(2, n)=1$ हैं। यदि $30\left(\begin{array}{l}30 \\ 0\end{array}\right)+29\left(\begin{array}{l}30 \\ 1\end{array}\right)+\ldots+2\left(\begin{array}{l}30 \\ 28\end{array}\right)+1\left(\begin{array}{l}30 \\ 29\end{array}\right)= n .2^{ m }$ हैं तो $n + m$ बराबर है I (यहाँ) $\left(\begin{array}{l} n \\ k \end{array}\right)={ }^{ n } C _{ k }$ है।

यदि ${ }^{20} C _{1}+\left(2^{2}\right){ }^{20} C _{2}+\left(3^{2}\right){ }^{20} C _{3}+\ldots \ldots+$ $\left(20^{2}\right)^{20} C _{20}= A \left(2^{\beta}\right)$, तो क्रमित युग्म $( A , \beta)$ बराबर है

$\frac{1}{{1!(n - 1)\,!}} + \frac{1}{{3!(n - 3)!}} + \frac{1}{{5!(n - 5)!}} + .... = $