Given that ${{T}_{r}}=\frac{{{r}^{2}}}{{{2}^{r}}}$, show that \({{T}_{r}}-{{T}_{r+1}}=\f

Question

Given that ${{T}_{r}}=\frac{{{r}^{2}}}{{{2}^{r}}}$, show that ${{T}_{r}}-{{T}_{r+1}}=\frac{{{\left( r-1 \right)}^{2}}}{{{2}^{r+1}}}-\frac{1}{{{2}^{r}}}$.[2] Hence, find $\sum\limits_{r=1}^{N}{\left( \frac{{{\left( r-1 \right)}^{2}}}{{{2}^{r+1}}}-\frac{1}{{{2}^{r}}} \right)}$ giving your answer in the form $\frac{1}{2}-\mathrm{f}\left( N \right)$.[2] Show that $\sum\limits_{r=1}^{N}{\frac{{{\left( r-1 \right)}^{2}}}{{{2}^{r+1}}}=\frac{3}{2}-\frac{{{\left( N+1 \right)}^{2}}}{{{2}^{N+1}}}-\frac{1}{{{2}^{N}}}}$.[2] Deduce an expression for $\sum\limits_{r=1}^{N-1}{\frac{{{r}^{2}}}{{{2}^{r}}}}$ in terms of $N$.[3]

Accepted Answer

(ii) $\frac{1}{2}-\frac{{{\left( N+1 \right)}^{2}}}{{{2}^{N+1}}}$ (iv) $\sum\limits_{r=1}^{N-1}{\frac{{{r}^{2}}}{{{2}^{r}}}=6-\frac{{{\left( N+1 \right)}^{2}}}{{{2}^{N-1}}}-\frac{1}{{{2}^{N-2}}}}$ since the first term is zero

2020 TJC P1 Q6

2020 TJC P1 Q6

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