જો {(1 + x)^{2016}} + x{(1 + x)^{2015}} + {x^ | vedclass

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Question

$S=(1+x)^{2016}+x(1+x)^{2015}+x^{2}(1+x)^{2014}$

$+\ldots+x^{2015}(1+x)+x^{2016}........(i)$

$\left(\frac{x}{1+x}\right) S=x(1+x)^{2015}+x^{2}(1

Vedclass · Accepted Answer

$\frac{{2017\,!\,}}{{17\,!\,2000\,!}}$

Vedclass · Answer

$S=(1+x)^{2016}+x(1+x)^{2015}+x^{2}(1+x)^{2014}$

$+\ldots+x^{2015}(1+x)+x^{2016}........(i)$

$\left(\frac{x}{1+x}ight) S=x(1+x)^{2015}+x^{2}(1+x)^{2014}$

$+\ldots +x^{2016}+\frac{x^{2017}}{1+x}........(ii)$

Subtracting $(i)$ from $(ii)$

$\frac{S}{1+x}=(1+x)^{2016}-\frac{x^{2017}}{1+x}$

$	herefore \quad S=(1+x)^{2017}-x^{2017}$

$a_{17}=$ coefficient of $x^{17}=^{2017} C_{17}$

$=\frac{2017 !}{17 ! 2000 !}$

જો ${(1 + x)^{2016}} + x{(1 + x)^{2015}} + {x^2}{(1 + x)^{2014}} + ....{x^{2016}} = \sum\limits_{i = 0}^{2016} {{a_i\,}{\,x^i}} $ જ્યાં $x\, \in \,R\,,\,x\, \ne \, - 1$ તો $a_{17}$ ની કિમત મેળવો.

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