compute-k-0-2k-1-2-2-k-1-

Question Number 17102 by tawa tawa last updated on 30/Jun/17

compute : \underset{k = 0}{\sum^{\infty}} \frac{2 k + 1}{2^{2 (k + 1)}}

Commented by prakash jain last updated on 01/Jul/17

S = \underset{k = 0}{\sum^{\infty}} \frac{2 k + 1}{2^{2 (k + 1)}}

S_{1} = \underset{k = 0}{\sum^{\infty}} \frac{2 k}{2^{2 (k + 1)}}

S_{2} = \underset{k = 0}{\sum^{\infty}} \frac{1}{2^{2 (k + 1)}} = \frac{1}{2^{2}} + \frac{1}{2^{4}} + \frac{1}{2^{6}} + \dots

S = S_{1} + S_{2}

S_{2} is infinite GP S_{2} = \frac{\frac{1}{2^{2}}}{1 - \frac{1}{2^{2}}} = \frac{1}{3}

S_{1} = \frac{2}{2^{2}} + \frac{3}{2^{4}} + \frac{4}{2^{6}} + \dots

\frac{S_{1}}{2^{2}} = \frac{2}{2^{4}} + \frac{3}{2^{6}} + \dots

S_{1} - \frac{S_{1}}{4} = \frac{2}{2^{2}} + \frac{1}{2^{4}} + \frac{1}{2^{6}} + \dots

S_{1} - \frac{S_{1}}{4} = \frac{1}{2^{2}} + (\frac{1}{2^{2}} + \frac{1}{2^{4}} + \frac{1}{2^{6}} + \dots)

\frac{3}{4} S_{1} = \frac{1}{2^{2}} + \frac{\frac{1}{2^{2}}}{1 - \frac{1}{2^{2}}} = \frac{1}{4} + \frac{1}{3} = \frac{7}{12}

S_{1} = \frac{7}{9}

S = S_{1} + S_{2} = \frac{7}{9} + \frac{1}{3} = \frac{10}{9}

Commented by ajfour last updated on 01/Jul/17

utter simple for you Sir .

Commented by tawa tawa last updated on 01/Jul/17

God bless you sir .