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\( \sum_{n=1}^{\infty} \frac{1}{(n+1)(n+2)(n+3) \ldots .(n+k)} \) is equal to |
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Answer» \(\sum\limits_{n=1}^{\infty}\frac{1}{(n+1)(n+2)(n+3)....(n+k)}\) \(=\frac{1}{2.3.4...(1+k)}+\frac{1}{3.4.5...(k+2)}+\frac{1}{4.5.6...(k+3)}+.....\infty\) terms (By expanding the sum) \(=\frac{1}{k-1}\left(\frac{k-1}{2.3.4...(1+k)}+\frac{k-1}{3.4.5...(k+2)}+\frac{1}{4.5.6...(k+3)}+....\infty \,terms\right)\) \(=\frac{1}{k-1}(\frac{(k+1)-2}{2.3.4...(1+k)}+\frac{(k+2)-3}{3.4.5...(k+2)}+\frac{(k+3)-4}{4.5.6...(k+3)}+....\infty\,terms)\) \(= \frac1{k-1}\bigg[\left(\frac1{2.3.4....k}-\frac1{3.4.5...(1+k)}\right)+\)\(\left(\frac{1}{3.4.5....(1+k)}-\frac1{3.5.6....(k+2)}\right)+\)\(\left(\frac1{4.5.6...(k+1)}-\frac1{5.6.7...(k+3)}\right)+....\infty\,terms\bigg]\) \(=\frac1{k-1}\times(\frac1{2.3.4...k}-0)\) (\(\because\) at infinite terms last term should be zero) \(=\frac1{k-1}\times\frac1{2.3.4...k}\) \(=\frac1{k!(k-1)}\) (\(\because\) 2.3.4...k = k!) ∑n=15n(n+1)(n+2)(n+3)1=3k21∑n=15(n(n+3)1−(n+1)(n+3)1)=3k∑n=15[31(n(n+3)3)−(n+1)(n+3)1]=32k∑n=15(3n1−n+11+n+21−3(n+3)1)=32k31(1+ |
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