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2020 TJC P1 Q6

2020 TJC P1 Q6

9 marks
Prelims
  1. 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]
  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]
  3. 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]
  4. Deduce an expression for \(\sum\limits_{r=1}^{N-1}{\frac{{{r}^{2}}}{{{2}^{r}}}}\) in terms of \(N\).[3]

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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

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