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Question Number 71776 by TawaTawa last updated on 19/Oct/19

Answered by tw000001 last updated on 22/Oct/19

$$\mathrm{Use}\mathrm{Harmonic}\mathrm{series}\mathrm{to}\mathrm{solve}.$$$$A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}$$$$=(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016})-2(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016})$$$$=(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016})-(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1008})$$$$=H_{2016}-H_{1008}$$$$B=\frac{2(\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2016})}{3025}$$$$=\frac{2(H_{2016}-H_{1008})}{3025}=\frac{2A}{3025}$$$$\therefore \frac{20A}{11B}=\frac{20\times 3025}{11\times 2}=2750$$

Commented by TawaTawa last updated on 22/Oct/19

$$\mathrm{Thanks}\mathrm{for}\mathrm{producing}\mathrm{more}\mathrm{steps},\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by TawaTawa last updated on 21/Oct/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{Thanks}\mathrm{for}\mathrm{your}\mathrm{time}$$

Commented by TawaTawa last updated on 21/Oct/19

$$\mathrm{Somebody}\mathrm{must}\mathrm{have}\mathrm{mistakenly}\mathrm{flag}\mathrm{instead}\mathrm{of}\mathrm{like}.$$$$\mathrm{Thanks}\mathrm{for}\mathrm{the}\mathrm{solution}\mathrm{sir}.$$

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