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  • Find the sum of series <img alt="\frac{1}{1 \cdot2 \cdot3}+\frac{1}{ 2\cdot3 \cdot4}+\frac{1}{3 \cdot4 \cdot 5}..... n\ terms \\" src="http://entrancecorner.oncodecogs.com/gif.latex?%5Cfrac%7B

Find the sum of series \frac{1}{1 \cdot2 \cdot3}+\frac{1}{ 2\cdot3 \cdot4}+\frac{1}{3 \cdot4 \cdot 5}..... n\ terms \\

Option: 1

[\frac{1}{2}-\frac{1}{(n+1)(n+2)}]


Option: 2

\frac{1}{2}[\frac{1}{2}-\frac{1}{(n)(n+1)}]


Option: 3

[\frac{1}{2}-\frac{1}{(n)(n+1)}]


Option: 4

\frac{1}{2}[\frac{1}{2}-\frac{1}{(n+1)(n+2)}]


Answers (1)

best_answer

\\S_{n}=\frac{1}{1 \cdot2 \cdot3}+\frac{1}{ 2\cdot3 \cdot4}+\frac{1}{3 \cdot4 \cdot 5}..... n\ terms \\ \\T_{r}=\frac{1}{r(r+1)(r+2)}\\\\\text{Last factor is (r+2) and first factor is r, their difference} (r+2)-r = 2\\\frac{(r+2)-r}{2}=1\\\text{Substituting this in }T_r\\T_{r}=\frac{1}{2} \cdot \frac{[r+2]-r}{r(r+1)(r+2)}\\ T_{r}=\frac{1}{2} \cdot [\frac{1}{r(r+1)}-\frac{1}{(r+1)(r+2)}]\\ S_{n}=\sum ^{n}_{1}T_{r}\\ S_{n}=\frac{1}{2}[(\frac{1}{1.2}-\frac{1}{2.3}) + (\frac{1}{2.3}-\frac{1}{3.4})+.......(\frac{1}{n.(n+1)}-\frac{1}{(n+1)(n+2)})]\\ S_{n}=\frac{1}{2}[\frac{1}{2}-\frac{1}{(n+1)(n+2)}]

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Divya Prakash Singh

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