A) \[e+1\]
B) \[e-1\]
C) \[{{e}^{-1}}\]
D) e
Correct Answer: D
Solution :
[d] \[\therefore \frac{2}{1}+\frac{4}{3}+\frac{6}{5}+......\infty \] \[=1+\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}.....to\,\infty \] \[={{e}^{1}}=e\] Hence, option [d] is correct.You need to login to perform this action.
You will be redirected in
3 sec