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Question Number 138295 by greg_ed last updated on 12/Apr/21

$$\mathbf{hi}!$$$$\mathbf{Simplify}:forn\in {\mathbb{N}}^{\ast },S_n=\underset{k=1}{\sum ^n}\frac{3k+8}{k(k+2)2^k}.$$

Answered by mr W last updated on 12/Apr/21

$$\frac{3k+8}{k(k+2)}=\frac{A}{k}+\frac{B}{k+2}=\frac{(A+B)k+2A}{k(k+2)}$$$$\Rightarrow 2A=8\Rightarrow A=4$$$$\Rightarrow A+B=3\Rightarrow B=-1$$$$S_n=\underset{k=1}{\sum ^n}\frac{3k+8}{k(k+2)2^k}$$$$=\underset{k=1}{\sum ^n}(\frac{4}{k}-\frac{1}{k+2})\frac{1}{2^k}$$$$=4\underset{k=1}{\sum ^n}\frac{1}{k{2}^k}-\underset{k=1}{\sum ^n}\frac{1}{(k+2)2^k}$$$$=4\underset{k=1}{\sum ^n}\frac{1}{k{2}^k}-4\underset{k=1}{\sum ^n}\frac{1}{(k+2)2^{k+2}}$$$$=4\underset{k=1}{\sum ^n}\frac{1}{k{2}^k}-4\underset{k=3}{\sum ^{n+2}}\frac{1}{k{2}^k}$$$$=\cancel{4\underset{k=1}{\sum ^n}\frac{1}{k{2}^k}}-\cancel{4\underset{k=1}{\sum ^n}\frac{1}{k{2}^k}}+4(\frac{1}{1\times 2^1}+\frac{1}{2\times 2^2}-\frac{1}{(n+1)2^{n+1}}-\frac{1}{(n+2)2^{n+2}})$$$$=\frac{5}{2}-\frac{3n+5}{(n+1)(n+2)2^n}$$

Commented by henderson last updated on 12/Apr/21

$$\mathrm{thank}\mathrm{u},\mathrm{sir}\mathrm{mr}\mathrm{W}!$$

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