Let be a sequence such that a _ 1 + a _ 2 + . . . + a _ n = n ^ 2 +...

Given,

${S}_{n} = \mathop{∑}_{i = 1}^{n}{a}_{i} = \dfrac{{n}^{2} + 3n}{\left(n + 1\right)\left(n + 2\right)}$

Now we know that,

${a}_{n} = {S}_{n} - {S}_{n - 1}$

$⇒ {a}_{n} = \dfrac{{n}^{2} + 3n}{\left(n + 1\right)\left(n + 2\right)} - \dfrac{{\left(n - 1\right)}^{2} + 3\left(n - 1\right)}{n\left(n + 1\right)}$

$⇒ {a}_{n} = \dfrac{4}{n\left(n + 1\right)\left(n + 2\right)}$

Now solving, $\mathop{∑}_{k = 1}^{10}\dfrac{1}{{a}_{k}} = \dfrac{1}{4}\mathop{∑}_{k = 1}^{10}k\left(k + 1\right)\left(k + 2\right)$

$⇒ \mathop{∑}_{k = 1}^{10}\dfrac{1}{{a}_{k}} = \dfrac{1}{16}\mathop{∑}_{k = 1}^{10}k\left(k + 1\right)\left(k + 2\right)\left(k + 3\right) - \left(k - 1\right)k\left(k + 1\right)\left(k + 2\right)$

$⇒ \mathop{∑}_{k = 1}^{10}\dfrac{1}{{a}_{k}} = \dfrac{1}{16}\left[\left(1 \cdot 2 \cdot 3 \cdot 4 . - 0\right) + \left(2 \cdot 3 \cdot 4 \cdot 5 - 1 \cdot 2 \cdot 3 \cdot 4\right) . . . . + \left(10 \cdot 11 \cdot 12 \cdot 13 - 9 \cdot 10 \cdot 11 \cdot 12\right)\right]$

$⇒ \mathop{∑}_{k = 1}^{10}\dfrac{1}{{a}_{k}} = \dfrac{1}{16}\left[10 \cdot 11 \cdot 12 \cdot 13 - 0\right]$

$⇒ \mathop{∑}_{k = 1}^{10}\dfrac{1}{{a}_{k}} = \dfrac{1}{2}\left(5 \cdot 11 \cdot 3 \cdot 13\right)$

$⇒ 28\mathop{∑}_{k = 1}^{10}\dfrac{1}{{a}_{k}} = \dfrac{28 \times 5 \times 11 \times 3 \times 13}{2}$

$⇒ 28\mathop{∑}_{k = 1}^{10}\dfrac{1}{{a}_{k}} = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$

Hence, there are six prime numbers in multiplications,

So, $m = 6$