Question Number 178530 by Acem last updated on 17/Oct/22
$${Conclude}\:{the}\:{value}\:{of}\:{the}\:{sum}: \\ $$$${S}_{{n}} =\begin{pmatrix}{{n}}\\{\mathrm{0}}\end{pmatrix}\:+\:\mathrm{2}\begin{pmatrix}{{n}}\\{\mathrm{1}}\end{pmatrix}+…+\:\mathrm{2}^{{n}} \begin{pmatrix}{{n}}\\{{n}}\end{pmatrix}\: \\ $$$$\:{with}\:{help}\:{of}\:\left(\mathrm{1}+\mathrm{2}{x}\right)^{{n}} \\ $$
Answered by Ar Brandon last updated on 17/Oct/22
$${S}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\overset{{n}} {\:}{C}_{{k}} \mathrm{2}^{{k}} =\left(\mathrm{2}+\mathrm{1}\right)^{{n}} =\mathrm{3}^{{n}} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\overset{{n}} {\:}{C}_{{k}} {a}^{{k}} {b}^{{n}−{k}} =\left({a}+{b}\right)^{{n}} \\ $$
Commented by Acem last updated on 17/Oct/22
$${Great}\:{Sir}! \\ $$