$\frac{10001 \times 100 !}{2 \times 1 !+5 \times 2 !+10 \times 3 !+\ldots+10001 \times 100 !}=$

Question

Aicharya · Accepted Answer

Here, general term of denominator is $\begin{aligned} & \sum_{n=1}^{100}\left(n^2+1\right) n !=\sum_{n=1}^{100}\left[(n+1)^2 n !-2 n \cdot n !\right] \ & =\sum_{n=1}^{100}[(n+1)(n+1) !-n \cdot n !]-\sum_{n=1}^{100} n \cdot n ! \ & =\sum_{n=1}^{100}[(n+1)(n+1) !-n \cdot n !]-\sum_{n=1}^{100}(n+1) !-n ! \ & {\left[\begin{array}{ccc} 2 \cdot 2 ! & - & 1 \cdot 1 ! \ +3 \cdot 3 ! & - & 2 \cdot 2 ! \ \vdots & & \ +101 \cdot 101 ! & -100 \cdot 100 ! \end{array}\right]-\left[\begin{array}{cc} 2 !- & 1 ! \ +3 !- & 2 ! \ \vdots \ +101 !-100 ! \end{array}\right]} \ & =[101 \cdot 101 !-1]-[101 !-1] \ \end{aligned}$ Denominator $=100 \cdot 101 !$ Consider, $\frac{10001 \times 100 !}{100 \times 101 \times 100 !}=\frac{10001}{10100}$ Hence, option (b) is correct.

\(\frac{10001 \times 100 !}{2 \times 1 !+5 \times 2 !+10 \times 3 !+\ldots+10001 \times 100 !}=\)

Solution