A) \[3\]
B) \[1\]
C) \[2\]
D) \[3/2\]
Correct Answer: C
Solution :
\[\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...,\] \[{{t}_{n}}=\frac{1}{1+2+3+...n}=\frac{2}{n(n+1)}=2\left[ \frac{1}{n}-\frac{1}{n+1} \right]\] \[\therefore \]Sum\[=S=2\left[ \left( \frac{1}{1}-\frac{1}{2} \right)+\left( \frac{1}{2}-\frac{1}{3} \right)+\left( \frac{1}{3}-\frac{1}{4} \right)+... \right]=2\]You need to login to perform this action.
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