Given that $\mathrm{f}\left( r \right)=r{{2}^{r}}$, show that \(\mathrm{f}\left( r+1 \ri

Question

Given that $\mathrm{f}\left( r \right)=r{{2}^{r}}$, show that $\mathrm{f}\left( r+1 \right)-\mathrm{f}\left( r \right)=\left( r+2 \right){{2}^{r}}$.[1] Hence show that $\sum\limits_{r=1}^{n}{\left( r+2 \right){{2}^{r}}=\left( n+1 \right){{2}^{n+1}}-2}$.[3] Use your answer in part (ii) to find $\sum\limits_{r=1}^{n}{r{{2}^{r}}}$. Give your answer in the form $\left( an+b \right){{2}^{n+1}}+c$, where $a$, $b$ and $c$ are integers to be determined.[3]

Accepted Answer

(ii) $\sum_{r=1}^n(r+1)2^r=(n+1)2^{n+1}-2$; (iii) $\sum_{r=1}^nr2^r=(n-1)2^{n+1}+2$, so $a=1,b=-1,c=2$

2020 VJC Promo Q3

2020 VJC Promo Q3

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