+526 Let \(S={1\over 2}+{3\over 2^2}+{5\over2^3}+...+{2n-1\over 2^n}\) ...(1)
⇒ \({S\over 2}={1\over 2^2}+{3\over 2^3}+{5\over2^4}+...+{2n-2\over 2^n}+{2n-1\over 2^{n+1}}\) ...(2)
Subtracting eq(2) from (1),
\({S\over2} = {1\over 2}+{2\over 2^2}+{2\over 2^3}+{2\over 2^4}+...+{1\over 2^n}-{2n-1\over 2^{n+1}}\)
\(={1\over 2}+({1\over 2}+{1\over 4}+{1\over 8}+....{1\over 2^n})-{2n-1\over 2^{n+1}}\)
\(={1\over2}+[{{1\over2}(1-{1\over2^n})\over 1-{1\over2}}]-{2n-1\over 2^{n+1}}\)
\(={1\over2}+1-{1\over 2^n}-{2n-1\over2^{n+1}}\)
\(={3\over 2}-[{2+2n-1\over 2^{n+1}}]\)
\({S\over 2} = {3\over 2}-{2n+1\over 2^{n+1}}\)
⇒\(S = 3-{2n+1\over 2^n}\)
+26367 Evaluate the sum \(\dfrac{1}{2} + \dfrac{3}{2^2} + \dfrac{5}{2^3} + \dfrac{7}{2^4} + \ldots\)
\(\begin{array}{|rcll|} \hline s &=& \dfrac{1}{2} &+ \dfrac{3}{2^2} + \dfrac{5}{2^3} + \dfrac{7}{2^4} + \ldots \\ s &=& 1*2^{-1} &+ 3*2^{-2} + 5*2^{-3} + 7*2^{-4} + 9*2^{-5} + \ldots \\ 2^{-1}s &=& &1*2^{-2} + 3*2^{-3} + 5*2^{-4} + 7*2^{-5} + \ldots \\ \hline s-2^{-1}s &=& 1*2^{-1} &+ 2 (\underbrace{2^{-2} + 2^{-3} + 2^{-4} + \ldots}_{=2^{-2}*\dfrac{1}{1-2^{-1}} } ) \\ s(1-2^{-1}) &=& 1*2^{-1} &+ 2*2^{-2}*\dfrac{1}{1-2^{-1}} \\ s*\dfrac{1}{2} &=& \dfrac{1}{2}&+\dfrac{1}{2}*\dfrac{1}{\dfrac{1}{2}} \\ s*\dfrac{1}{2} &=& \dfrac{1}{2}&+\dfrac{1}{2}*2 \\ s*\dfrac{1}{2} &=& \dfrac{1}{2}&*(1+2) \\ s &=& &1+2 \\ \mathbf{s} &=& \mathbf{3} \\ \hline \end{array}\)
