If $A =[a_{ij}]_n×n$, where $a_{ij}=i^{100}+j^{100}$, $\underset{n→∞}{\lim}\left(\frac{\sum\limits_{i=1}^{n}a_{ii}}{n^{101}}\right)$ equals
Answer & explanation
Correct answer: option 3
We have, $a_{ij}=i^{100}+j^{100}$
$⇒a_{ii}=2i^{100}$
$∴\underset{n→∞}{\lim}\left(\frac{\sum\limits_{i=1}^{n}a_{ij}}{n^{101}}\right)=\underset{n→∞}{\lim}\left(\frac{\sum\limits_{i=1}^{n}(2i^{100})}{n^{101}}\right)$
$=\underset{n→∞}{2\lim}\sum\limits_{i=1}^{n}(\frac{i}{n})^{100}.\frac{1}{n}=2\int\limits_0^1x^{100}dx=\frac{2}{101}$