A) \[\frac{n}{n+1}\]
B) \[\frac{2n}{n+1}\]
C) \[\frac{2}{n\,(n+1)}\]
D) \[\frac{2\,(n+1)}{n+2}\]
Correct Answer: D
Solution :
\[{{T}_{n}}=\frac{1}{\left[ \frac{n(n+1)}{2} \right]}=2\left[ \frac{1}{n}-\frac{1}{n+1} \right]\] Put \[n=1,\,2,\,3,....,(n+1)\] \[{{T}_{1}}=2\,\left[ \frac{1}{1}-\frac{1}{2} \right]\,,\,\,{{T}_{2}}=2\,\left[ \frac{1}{2}-\frac{1}{3} \right]\,,........,\] \[{{T}_{n+1}}=2\left[ \frac{1}{n+1}-\frac{1}{n+2} \right]\] Hence sum of (n + 1) terms \[=\sum\limits_{k=1}^{n+1}{{{T}_{k}}}\] \[\Rightarrow {{S}_{n+1}}=2\left[ 1-\frac{1}{n+2} \right]\]\[\Rightarrow {{S}_{n+1\,}}\,=\frac{2(n+1)}{(n+2)}\].
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