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Q.
The value of $\left(\displaystyle\sum_{k=0}^{100} \frac{k+3}{(k+1) !+(k+2) !+(k+3) !}\right)+\frac{1}{(103) !}$ is equal to
Binomial Theorem
Solution:
$\displaystyle\sum_{k=0}^{100}\left(\frac{k+3}{(k+1) !+(k+2) !+(k+3) !}\right)=\displaystyle\sum_{k=0}^{100} \frac{(k+3)}{(k+1) ![1+k+2+(k+3)(k+2)]}$
$=\displaystyle\sum_{ k =0}^{100} \frac{ k +3}{( k +1) !\left( k ^2+6 k +9\right)}=\displaystyle\sum \frac{ k +3}{( k +1)( k +3)^2}=\sum_{ k =0}^{100} \frac{1}{( k +3)( k +1) !} $
$=\displaystyle\sum_{ k =0}^{100} \frac{ k +2=( k +3)-1}{( k +3) !}=\displaystyle\sum_{ k =0}^{100} \frac{1}{( k +2) !}-\frac{1}{( k +3) !}=\frac{1}{2 !}-\frac{1}{(103) !} $
$\therefore S =\frac{1}{2} \Rightarrow \text { B}$