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Sum-up-the-following-to-nth-term-5-3-10-8-17-15-26-24-

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Question Number 106515 by nimnim last updated on 05/Aug/20
Sum up the following to nth term: (5/3)+((10)/8)+((17)/(15))+((26)/(24))+........
$${Sum}\:{up}\:{the}\:{following}\:{to}\:{nth}\:{term}: \\ $$$$\:\:\:\:\:\:\:\:\:\frac{\mathrm{5}}{\mathrm{3}}+\frac{\mathrm{10}}{\mathrm{8}}+\frac{\mathrm{17}}{\mathrm{15}}+\frac{\mathrm{26}}{\mathrm{24}}+…….. \\ $$
Answered by Dwaipayan Shikari last updated on 05/Aug/20
S=1+(2/3)+1+(2/8)+1+(2/(15))+1+(2/(24))+.. S=(1+1+1+...n)+2((1/3)+(1/8)+(1/(15))+...) S=n+2Σ_(n=1) ^n (1/(n(n+2))) S=n+2Σ_(n=1) ^n ((n+1)/(n(n+2)(n+1))) S=n+2Σ^n (1/((n+1)(n+2)))+(1/(n(n+2)(n+1))) S=n+2Σ^n (1/(n+1))−(1/(n+2))+Σ^n ((n+2−n)/(n(n+2)(n+1))) S=n+2((1/2)−(1/(n+2)))+Σ^n (1/(n(n+1)))−Σ^n (1/((n+1)(n+2))) S=n+(1−(2/(n+2)))+(1−(1/(n+1)))−Σ^n (1/(n+1))−(1/(n+2)) S=n+(1−(2/(n+2)))+(1−(1/(n+1)))−(1/2)+(1/(n+2)) S=n+(3/2)−(1/(n+2))−(1/(n+1))
$${S}=\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}}+\mathrm{1}+\frac{\mathrm{2}}{\mathrm{8}}+\mathrm{1}+\frac{\mathrm{2}}{\mathrm{15}}+\mathrm{1}+\frac{\mathrm{2}}{\mathrm{24}}+.. \\ $$$${S}=\left(\mathrm{1}+\mathrm{1}+\mathrm{1}+…{n}\right)+\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{15}}+…\right) \\ $$$${S}={n}+\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{n}\left({n}+\mathrm{2}\right)} \\ $$$${S}={n}+\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{n}+\mathrm{1}}{{n}\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right)} \\ $$$${S}={n}+\mathrm{2}\overset{{n}} {\sum}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}+\frac{\mathrm{1}}{{n}\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right)} \\ $$$${S}={n}+\mathrm{2}\overset{{n}} {\sum}\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{2}}+\overset{{n}} {\sum}\frac{{n}+\mathrm{2}−{n}}{{n}\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right)} \\ $$$${S}={n}+\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{{n}+\mathrm{2}}\right)+\overset{{n}} {\sum}\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}−\overset{{n}} {\sum}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)} \\ $$$${S}={n}+\left(\mathrm{1}−\frac{\mathrm{2}}{{n}+\mathrm{2}}\right)+\left(\mathrm{1}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)−\overset{{n}} {\sum}\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{2}} \\ $$$${S}={n}+\left(\mathrm{1}−\frac{\mathrm{2}}{{n}+\mathrm{2}}\right)+\left(\mathrm{1}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{{n}+\mathrm{2}} \\ $$$${S}={n}+\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{1}}{{n}+\mathrm{2}}−\frac{\mathrm{1}}{{n}+\mathrm{1}} \\ $$$$ \\ $$
Commented by nimnim last updated on 06/Aug/20
Thank you Sir.
$${Thank}\:{you}\:{Sir}. \\ $$

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