Silence Of The Sum

$T_{n} = \frac{1}{2^{r}}. \frac{(r + 1)^{2} + 1}{r(r + 1)}$

$=\frac{1}{2^{r}}. \frac{(r + 1)^{2} -1 + 2}{r(r + 1)}$

$= \frac{1}{2^{r}} (\frac{r(r + 2) }{r(r + 1)} + \frac{2}{r(r + 1)})$

$= \frac{1}{2^{r}} (\frac{r }{(r + 1)} + \frac{2}{r + 1} + \frac{2}{r} - \frac{2}{r + 1})$

$= \frac{1}{2^{r}} (\frac{r + 1 - 1 }{(r + 1)} + \frac{2}{r + 1} + \frac{2}{r} - \frac{2}{r + 1})$

$= \frac{1}{2^{r}} ( 1 - \frac{1 }{(r + 1)} + \frac{2}{r + 1} + \frac{2}{r} - \frac{2}{r + 1})$

$= \frac{1}{2^{r}} + \frac{1}{2^r}(\frac{2}{r} - \frac{1}{r + 1})$

adding all

we get

$= \frac{1}{2} \frac{2}{1} + \frac{1}{2}(\frac{2}{1} - \frac{1}{2}) + \frac{1}{2^{2}}(\frac{2}{2} - \frac{1}{3}) + \frac{1}{2^{3}}(\frac{2}{3} - \frac{1}{4}) + ...........................$

$1 + 1 - \frac{1}{2.2} + \frac{1}{2.2} ......................$

$= 2$

OR Simply open the square in the numerator and do the above you will get the answer

It is a simple summation of rth term 2^(-1) { 1 + (2/r) - (1/(r + 1))} , giving 2 sum as plus minus terms cancel in next term. Check by putting values for r.

$S=\sum _{ i=1 }^{ \infty }{ \frac { 1 }{ { 2 }^{ i } } } +\sum _{ j=1 }^{ \infty }{ \frac { 1 }{ j{ 2 }^{ j-1 } } -\frac { 1 }{ (j+1){ 2 }^{ j } } } =1+1=2\quad$ Note the second summation is converging as it is of the form f(k)-f(k+1)