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Question Number 197880 by universe last updated on 02/Oct/23

find the sum of infinite series

\frac{1}{2^{1}} \cdot \frac{1}{3^{2}} + \frac{1}{2^{2}} \cdot \frac{1}{3^{4}} (1^{2} + 2^{2} + 3^{2}) + \frac{1}{2^{3}} \cdot \frac{1}{3^{6}} (1^{2} + 2^{2} + 3^{2} + \dots + 7^{2}) +

\frac{1}{2^{4}} \cdot \frac{1}{3^{8}} (1^{2} + 2^{2} + 3^{2} + \dots + 15^{2}) + \dots \dots . .

Answered by qaz last updated on 02/Oct/23

\underset{i = 1}{\sum^{\infty}} \frac{1}{18^{i}} (\underset{j = 1}{\sum^{2^{i} - 1}} j^{2}) = \frac{1}{6} \underset{i = 1}{\sum^{\infty}} \frac{1}{18^{i}} \cdot 2^{i} (2^{i} - 1) (2^{i + 1} - 1)

= \frac{1}{6} \underset{i = 1}{\sum^{\infty}} (2 \cdot {(\frac{4}{9})}^{i} - 3 \cdot {(\frac{2}{9})}^{i} + \frac{1}{9^{i}})

= \frac{1}{6} (2 \cdot \frac{\frac{4}{9}}{1 - \frac{4}{9}} - 3 \cdot \frac{\frac{2}{9}}{1 - \frac{2}{9}} + \frac{\frac{1}{9}}{1 - \frac{1}{9}}) = \frac{243}{1680}

Commented by universe last updated on 03/Oct/23

thank you sir