Let < a n > be a sequence such that a 1 + a 2 + . . . + a n = n 2 + 3 n n + 1 n + 2 . If 28 ∑…

Given,

$S_n=\sum _{i=1}^n{a}_i=\frac{{\displaystyle n^2+3n}}{{\displaystyle \left(n+1\right)\left(n+2\right)}}$

Now we know that,

$a_n=S_n-S_{n-1}$

$\Rightarrow a_n=\frac{n^2+3n}{\left(n+1\right)\left(n+2\right)}-\frac{{\left(n-1\right)}^2+3\left(n-1\right)}{n\left(n+1\right)}$

$\Rightarrow a_n=\frac{4}{n\left(n+1\right)\left(n+2\right)}$

Now solving, $\sum _{k=1}^{10}\frac{1}{a_k}=\frac{1}{4}\sum _{k=1}^{10}k\left(k+1\right)\left(k+2\right)$

$\Rightarrow \sum _{k=1}^{10}\frac{1}{a_k}=\frac{1}{16}\sum _{k=1}^{10}k\left(k+1\right)\left(k+2\right)\left(k+3\right)-\left(k-1\right)k\left(k+1\right)\left(k+2\right)$

$\Rightarrow \sum _{k=1}^{10}\frac{1}{a_k}=\frac{1}{16}\left[\left(1·2·3·4.-0\right)+\left(2·3·4·5-1·2·3·4\right)....+\left(10·11·12·13-9·10·11·12\right)\right]$

$\Rightarrow \sum _{k=1}^{10}\frac{1}{a_k}=\frac{1}{16}\left[10·11·12·13-0\right]$

$\Rightarrow \sum _{k=1}^{10}\frac{1}{a_k}=\frac{1}{2}\left(5·11·3·13\right)$

$\Rightarrow 28\sum _{k=1}^{10}\frac{1}{a_k}=\frac{28\times 5\times 11\times 3\times 13}{2}$

$\Rightarrow 28\sum _{k=1}^{10}\frac{1}{a_k}=2·3·5·7·11·13$

Hence, there are six prime numbers in multiplications,

So, $m=6$