Are you good at summation

Here's somewhat of a generalization to the problem:

Consider a generalized version of this sum as $\mathcal{M}$ where $a,b,\delta_1,\delta_2\in\Bbb{Z}$ such that $\delta_1\neq \delta_2$ .

Also, consider that $(n+i)\neq 0~\forall~i,n\mid \delta_1\leq i\leq \delta_2~\land~a\leq n\leq b$ . Then,

$\large\mathcal{M}=\sum_{n=a}^b\left(\frac{1}{\displaystyle\prod_{i=\delta_1}^{\delta_2}(n+i)}\right)=\frac{1}{\delta_2-\delta_1}\cdot\sum_{n=a}^b\left(\frac{1}{\displaystyle\prod_{i=\delta_1}^{\delta_2-1}(n+i)}-\frac{1}{\displaystyle\prod_{i=\delta_1+1}^{\delta_2}(n+i)}\right)$

By telescopy, the sum evaluates to,

$\large\mathcal{M}=\frac{1}{\delta_2-\delta_1}\left(\frac{1}{\displaystyle\prod_{i=\delta_1}^{\delta_2-1}(a+i)}-\frac{1}{\displaystyle\prod_{i=\delta_1+1}^{\delta_2}(b+i)}\right)=\frac{1}{\delta_2-\delta_1}\left(\frac{1}{\displaystyle\prod_{i=a+\delta_1}^{a+\delta_2-1}(i)}-\frac{1}{\displaystyle\prod_{i=b+\delta_1+1}^{b+\delta_2}(i)}\right)$

When factorial notation is defined, it can be rewritten as,

$\mathcal{M}=\frac{1}{\delta_2-\delta_1}\left(\frac{(a+\delta_1-1)!}{(a+\delta_2-1)!}-\frac{(b+\delta_1)!}{(b+\delta_2)!}\right)$

Otherwise, we simply use the normal form (product form) obtained earlier instead of the factorial notation.

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For the problem here, plug in $a=1,~b=25,~\delta_1=0,~\delta_2=5$ which satisfies all of the presumed conditions of our generalization. So, it gives, for the problem here,

$\mathcal{M}=\frac{1}{5-0}\left(\frac{0!}{5!}-\frac{25!}{30!}\right)=\frac{28501}{17100720}$

$M = \sum_{n=1}^{25}\dfrac{1}{(n)(n+1)(n+2)(n+3)(n+4)(n+5)}$

The given summation can be written as telescopic :

$M =\color{#D61F06}{\frac{1}{5}}\displaystyle \sum_{n=1}^{25} \frac{1}{(n)(n+1)(n+2)(n+3)(n+4)} - \frac{1}{(n+1)(n+2)(n+3)(n+4)(n+5)} \\ 5M= \frac{1}{(1)(2)(3)(4)(5)} - \frac{1}{(26)(27)(28)(29)(30)} \\ 5M= \dfrac{1}{120} - \frac{1}{17100720} \\ \Rightarrow M = \frac{28501}{17100720}$

Hence , $a+b = 28501+17100720 = \Large\boxed{17129221}$ .

Write $5M = \frac{1}{(n)(n+1)(n+2)(n+3)(n+4)}-\frac{1}{(n+1)(n+2)(n+3)(n+4)(n+5)}$ . When we write few terms, terms began to cancel each other out and we get $5M=\frac{1}{120}-\frac{1}{(26)(27)(28)(29)(30)}$ And $M=\frac{28501}{17100720}$ Therefore $\boxed{A+B =17129221}$ .