Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

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.

$${Thank}\:{you}\:{Sir}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com