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  4. (10001 × 100 !/2 × 1 !+5 × 2 !+10 × 3 !+ ldots+10001 × 100 !)=

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

AP EAMCETAP EAMCET 2020

Solution:

Here, general term of denominator is
$\displaystyle\sum_{n=1}^{100}\left(n^{2}+1\right) n !=\displaystyle\sum_{n=1}^{100}\left[(n+1)^{2} n !-2 n, n !\right]$
$=\displaystyle\sum_{n=1}^{100}[(n+1)(n+1) !-n \cdot n !]-\sum_{n=1}^{100} n \cdot n !$
$=\displaystyle\sum_{n=1}^{100}[(n+1)(n+1) !-n \cdot n !]-\sum_{n=1}^{100}(n+1) !-n !$
$\begin{bmatrix}2 \cdot 2 ! & - & 1 \cdot 1 ! \\ +3 \cdot 3 ! & - & 2 \cdot 2 ! \\ \vdots & & \\ +101 \cdot 101 ! & -&100 \cdot 100 !\end{bmatrix}-\begin{bmatrix}2 ! & -&1 ! \\ +3 ! & - & 2 ! \\ \vdots & & \\ + 101 ! & - & 100 !\end{bmatrix}$
$=[101 \cdot 101 !-1]-[101 !-1]$
Denominator $=100 \cdot 101 !$
Consider, $\frac{10001 \times 100 !}{100 \times 101 \times 100 !}$
$=\frac{10001}{10100}$