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Question Number 121949 by liberty last updated on 12/Nov/20

Evaluate the sum \frac{2}{3 + 1} + \frac{2^{2}}{3^{2} + 1} + \frac{2^{3}}{3^{4} + 1} + \dots + \frac{2^{n + 1}}{3^{2^{n}} + 1} .

Answered by bobhans last updated on 12/Nov/20

c o n s i d e r \frac{1}{a + 1} = \frac{a - 1}{a^{2} + 1} = \frac{a}{a^{2} - 1} - \frac{1}{a^{2} - 1}

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

w h i c h i m p l i e s \frac{1}{2} . \frac{1}{a + 1} = \frac{1}{2 (a - 1)} - \frac{1}{a^{2} - 1}

s u b s t i t u t e i d e n t i t y w i t h a = 3^{2^{k}}

w e o b t a i n \frac{1}{2 (3^{2^{k}} + 1)} = \frac{1}{2 (3^{2^{k}} - 1)} - \frac{1}{3^{2^{k + 1}} - 1}

m u l t i p l y i n g t h i s b y 2^{k + 2}, w e g e t

t h e r e l a t i o n \frac{2^{k + 1}}{3^{2^{k}} + 1} = \frac{2^{k + 1}}{3^{2^{k}} - 1} - \frac{2^{k + 2}}{3^{2^{k + 1}} - 1}

T h u s

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

\underset{k = 0}{\sum^{n}} (\frac{2^{k + 1}}{3^{2^{k}} - 1} - \frac{2^{k + 2}}{3^{2^{k + 1}} - 1})

= 1 - \frac{2^{n + 2}}{3^{2^{n + 1}} - 1} .

Commented by liberty last updated on 12/Nov/20

good \dots .

Commented by sewak last updated on 17/Nov/20

Quite good

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