A) \[{{a}_{n+1}}>{{a}_{n}}\]
B) \[{{a}_{n+1}}<{{a}_{n}}\]
C) \[{{a}_{n+1}}={{a}_{n}}\]
D) \[{{a}_{n+1}}-{{a}_{n}}=1/n\]
Correct Answer: B
Solution :
[b] We have \[{{a}_{n}}=\frac{1}{n+1}\sum\limits_{k=1}^{n}{\left( \frac{1}{k}+\frac{1}{n+1-k} \right)}\] \[=\frac{2}{n+1}\sum\limits_{k=1}^{n}{\frac{1}{k}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \because \,\,\,\sum\limits_{k=1}^{n}{\frac{1}{k}=\sum\limits_{k=1}^{n}{\frac{1}{n+1-k}}} \right)\] For \[n\ge 2,\] \[\frac{1}{2}({{a}_{n}}-{{a}_{n}}+1)=\frac{1}{n+1}\sum\limits_{k=1}^{n}{\frac{1}{k}}-\frac{1}{n+2}\sum\limits_{k=1}^{n+1}{\frac{1}{k}}\] \[=\left( \frac{1}{n+1}-\frac{1}{n+2} \right)\sum\limits_{k=1}^{n}{\frac{1}{k}}-\frac{1}{(n+1)(n+2)}\] \[=\frac{1}{(n+1)(n+2)}\sum\limits_{k=2}^{n}{\frac{1}{k}}>0\] \[\Rightarrow \,\,\,\,\,\,{{a}_{n}}>{{a}_{n+1}}\]
You need to login to perform this action.
You will be redirected in
3 sec
