A) \[\frac{1}{3n+1}\]
B) \[\frac{1}{{{(3n+1)}^{1/2}}}\]
C) \[\frac{1}{{{(3n+1)}^{2}}}\]
D) \[\frac{1}{{{(3n+1)}^{1/2}}}\]
Correct Answer: B
Solution :
| [b] By setting\[n=1,\]we get \[\frac{2n!}{{{2}^{2n}}.{{(n!)}^{2}}}\]as \[\frac{2}{{{2}^{2}}.{{(1)}^{2}}}=\frac{2}{4}=\frac{1}{2}.\] |
| Second alternative gives \[\frac{1}{2}\]. |
| Upon setting \[n=2,\frac{2n!}{{{2}^{2n}}.{{(n)}^{2}}}\]becomes \[\frac{3}{8}\], second alternative |
| Becomes \[\frac{1}{\sqrt{7}}\]which is greater than \[\frac{3}{8}.\] |
You need to login to perform this action.
You will be redirected in
3 sec
