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

Question

Given that $\mathrm{f}\left( r \right)=\frac{{{r}^{2}}}{{{2}^{r}}}$, show that $\mathrm{f}\left( r \right)-\mathrm{f}\left( r+1 \right)=\frac{{{\left( r-1 \right)}^{2}}}{{{2}^{r+1}}}-\frac{1}{{{2}^{r}}}$. [1] Hence, find $\sum\limits_{r=1}^{n}{\left[ \frac{{{\left( r-1 \right)}^{2}}}{{{2}^{r+1}}}-\frac{1}{{{2}^{r}}} \right]}$. [3] Deduce $\sum\limits_{r=1}^{n}{\left[ \frac{{{\left( r-1 \right)}^{2}}}{{{2}^{r+1}}} \right]}$. [3] Show that $\sum\limits_{r=1}^{n}{\frac{{{r}^{2}}}{{{2}^{r}}}=6-\frac{{{n}^{2}}+4n+6}{{{2}^{n}}}}$. [3]

Accepted Answer

$\displaystyle\sum_{r=1}^{n}\frac{r^2}{2^r} = 6 - \frac{n^2 + 4n + 6}{2^n}$

2007 NYJC P1 Q8 [Modified]

2007 NYJC P1 Q8 [Modified]

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