Question Number 151097 by naka3546 last updated on 18/Aug/21
$${Find}\:\:{sum}\:\:{of}\:\:{this}\:\:{expression}\:. \\ $$$$\begin{pmatrix}{{n}}\\{\mathrm{1}}\end{pmatrix}\:+\:\mathrm{2}\begin{pmatrix}{{n}}\\{\mathrm{2}}\end{pmatrix}\:+\:\mathrm{3}\begin{pmatrix}{{n}}\\{\mathrm{3}}\end{pmatrix}\:+\:\mathrm{4}\begin{pmatrix}{{n}}\\{\mathrm{4}}\end{pmatrix}\:+\:\ldots\:+\:{n}\begin{pmatrix}{{n}}\\{{n}}\end{pmatrix} \\ $$$${Please}\:\:{show}\:\:{your}\:\:{workings}.\:{Thank}\:\:{you}\:. \\ $$
Answered by Olaf_Thorendsen last updated on 18/Aug/21
$$\mathrm{S}_{{n}} \:=\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}\mathrm{C}_{{k}} ^{{n}} \\ $$$${f}\left({x}\right)\:=\:\left({x}+\mathrm{1}\right)^{{n}} \:=\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{C}_{{k}} ^{{n}} {x}^{{k}} \\ $$$${f}'\left({x}\right)\:=\:{n}\left({x}+\mathrm{1}\right)^{{n}−\mathrm{1}} \:=\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}\mathrm{C}_{{k}} ^{{n}} {x}^{{k}−\mathrm{1}} \\ $$$${f}'\left(\mathrm{1}\right)\:=\:{n}\mathrm{2}^{{n}−\mathrm{1}} \:=\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}\mathrm{C}_{{k}} ^{{n}} \:=\:\mathrm{S}_{{n}} \\ $$
Answered by mr W last updated on 18/Aug/21
![S_n = ((n),(1) ) + 2 ((n),(2) ) + 3 ((n),(3) ) + 4 ((n),(4) ) + … + n ((n),(n) ) S_n =0 ((n),(0) )+ ((n),(1) ) + 2 ((n),(2) ) + 3 ((n),(3) ) + 4 ((n),(4) ) + … + n ((n),(n) ) S_n =n ((n),(n) )+(n−1) ((n),((n−1)) ) +(n− 2) ((n),((n−2)) ) +(n− 3) ((n),((n−3)) ) + … + 0 ((n),((n−n)) ) S_n =n ((n),(0) )+(n−1) ((n),(1) ) +(n− 2) ((n),(2) ) +(n− 3) ((n),(3) ) + … + 0 ((n),(n) ) 2S_n =n[ ((n),(0) )+ ((n),(1) ) + ((n),(2) ) + ((n),(3) ) + … + ((n),(n) )] 2S_n =n×2^n ⇒S_n =n×2^(n−1)](Q151129.png)
$${S}_{{n}} =\begin{pmatrix}{{n}}\\{\mathrm{1}}\end{pmatrix}\:+\:\mathrm{2}\begin{pmatrix}{{n}}\\{\mathrm{2}}\end{pmatrix}\:+\:\mathrm{3}\begin{pmatrix}{{n}}\\{\mathrm{3}}\end{pmatrix}\:+\:\mathrm{4}\begin{pmatrix}{{n}}\\{\mathrm{4}}\end{pmatrix}\:+\:\ldots\:+\:{n}\begin{pmatrix}{{n}}\\{{n}}\end{pmatrix} \\ $$$${S}_{{n}} =\mathrm{0}\begin{pmatrix}{{n}}\\{\mathrm{0}}\end{pmatrix}+\begin{pmatrix}{{n}}\\{\mathrm{1}}\end{pmatrix}\:+\:\mathrm{2}\begin{pmatrix}{{n}}\\{\mathrm{2}}\end{pmatrix}\:+\:\mathrm{3}\begin{pmatrix}{{n}}\\{\mathrm{3}}\end{pmatrix}\:+\:\mathrm{4}\begin{pmatrix}{{n}}\\{\mathrm{4}}\end{pmatrix}\:+\:\ldots\:+\:{n}\begin{pmatrix}{{n}}\\{{n}}\end{pmatrix} \\ $$$${S}_{{n}} ={n}\begin{pmatrix}{{n}}\\{{n}}\end{pmatrix}+\left({n}−\mathrm{1}\right)\begin{pmatrix}{{n}}\\{{n}−\mathrm{1}}\end{pmatrix}\:+\left({n}−\:\mathrm{2}\right)\begin{pmatrix}{{n}}\\{{n}−\mathrm{2}}\end{pmatrix}\:+\left({n}−\:\mathrm{3}\right)\begin{pmatrix}{{n}}\\{{n}−\mathrm{3}}\end{pmatrix}\:+\:\ldots\:+\:\mathrm{0}\begin{pmatrix}{{n}}\\{{n}−{n}}\end{pmatrix} \\ $$$${S}_{{n}} ={n}\begin{pmatrix}{{n}}\\{\mathrm{0}}\end{pmatrix}+\left({n}−\mathrm{1}\right)\begin{pmatrix}{{n}}\\{\mathrm{1}}\end{pmatrix}\:+\left({n}−\:\mathrm{2}\right)\begin{pmatrix}{{n}}\\{\mathrm{2}}\end{pmatrix}\:+\left({n}−\:\mathrm{3}\right)\begin{pmatrix}{{n}}\\{\mathrm{3}}\end{pmatrix}\:+\:\ldots\:+\:\mathrm{0}\begin{pmatrix}{{n}}\\{{n}}\end{pmatrix} \\ $$$$\mathrm{2}{S}_{{n}} ={n}\left[\begin{pmatrix}{{n}}\\{\mathrm{0}}\end{pmatrix}+\begin{pmatrix}{{n}}\\{\mathrm{1}}\end{pmatrix}\:+\begin{pmatrix}{{n}}\\{\mathrm{2}}\end{pmatrix}\:+\begin{pmatrix}{{n}}\\{\mathrm{3}}\end{pmatrix}\:+\:\ldots\:+\:\begin{pmatrix}{{n}}\\{{n}}\end{pmatrix}\right] \\ $$$$\mathrm{2}{S}_{{n}} ={n}×\mathrm{2}^{{n}} \\ $$$$\Rightarrow{S}_{{n}} ={n}×\mathrm{2}^{{n}−\mathrm{1}} \\ $$