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Question Number 13121 by tawa tawa last updated on 14/May/17

$$\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}:\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+...+\frac{2}{9999}$$

Answered by nume1114 last updated on 15/May/17

$$\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+...+\frac{2}{9999}$$$$=\frac{2}{3\times 5}+\frac{2}{5\times 7}+\frac{2}{7\times 9}+...+\frac{2}{99\times 101}$$$$=\underset{k=1}{\sum ^{49}}\frac{2}{(2k+1)(2k+3)}$$$$=\underset{k=1}{\sum ^{49}}(\frac{1}{2k+1}-\frac{1}{2k+3})$$$$=(\frac{1}{3}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{9})+...+(\frac{1}{99}-\frac{1}{101})$$$$=\frac{1}{3}-\frac{1}{101}$$$$=\frac{98}{303}$$

Commented by tawa tawa last updated on 15/May/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by nume1114 last updated on 15/May/17

$$oh,youarerightsir.I^{\prime }vecorrectedit.$$

Commented by RasheedSindhi last updated on 15/May/17

$$\mathcal{N}i\mathcal{CE}!$$

Commented by tawa tawa last updated on 15/May/17

$$\mathrm{but}\mathrm{the}\mathrm{answer}=\frac{98}{303}$$$$\mathrm{mistake}\mathrm{in}\mathrm{your}\mathrm{L}.\mathrm{C}.\mathrm{M}$$

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