By considering $\frac{1}{1+{{2}^{r}}}-\frac{1}{1+{{2}^{r+1}}}$, show that $\sum\limits_{r=1}^{N}{\fr

Question

By considering $\frac{1}{1+{{2}^{r}}}-\frac{1}{1+{{2}^{r+1}}}$, show that $\sum\limits_{r=1}^{N}{\frac{{{2}^{r}}}{\left( 1+{{2}^{r}} \right)\left( 1+{{2}^{r+1}} \right)}}=\frac{1}{3}-\frac{1}{1+{{2}^{n+1}}}$. [4] Hence, find $\sum\limits_{r=1}^{N}{\frac{{{2}^{r+2}}}{\left( 1+{{2}^{r+2}} \right)\left( 1+{{2}^{r+3}} \right)}}$. [3]

Accepted Answer

$\dfrac{1}{9}-\dfrac{1}{1+2^{N+3}}$

2018 VJC Promo Q4

2018 VJC Promo Q4

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