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Question Number 203750 by patrice last updated on 27/Jan/24

Answered by witcher3 last updated on 27/Jan/24

$$\frac{4(\mathrm{k}+2)-\mathrm{k}}{\mathrm{k}(\mathrm{k}+2)2^{\mathrm{k}}}=\frac{4}{\mathrm{k}.2^{\mathrm{k}}}-\frac{1}{(\mathrm{k}+2)2^{\mathrm{k}}}=\frac{1}{\mathrm{k}.2^{\mathrm{k}-2}}-\frac{1}{(\mathrm{k}+2)2^{\mathrm{k}}}={\mathrm{V}}_{\mathrm{k}}-{\mathrm{V}}_{\mathrm{k}+2}$$$${\mathrm{s}}_{\mathrm{n}}=\mathrm{\Sigma }({\mathrm{V}}_{\mathrm{k}}-{\mathrm{V}}_{\mathrm{k}+2})={\mathrm{V}}_1+{\mathrm{V}}_2-{\mathrm{V}}_{\mathrm{n}+2}-{\mathrm{V}}_{\mathrm{n}+1}$$$$=\frac{5}{2}-\frac{1}{(\mathrm{n}+2)2^{\mathrm{n}}}-\frac{1}{(\mathrm{n}+1)2^{\mathrm{n}-1}}$$

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