Let α = 1 ^ 2 + 4 ^ 2 + 8 ^ 2 + 13 ^ 2 + 19 ^ 2 + 26 ^ 2 + . upto...

Given,

$\alpha = {1}^{2} + {4}^{2} + {8}^{2} \ldots$

Now, let $\sqrt{{t}_{n}} = a{n}^{2} + bn + c$

$1 = a + b + c . . . . \left(i\right)$

$4 = 4a + 2b + c . . . . \left(ii\right)$

$8 = 9a + 3b + c . . . . \left(iii\right)$

Now, solving equation \left(i\right), \left(ii\right) \text{ and } \left(iii\right)$we get,</p><p>$a = \dfrac{1}{2}, b = \dfrac{3}{2}, c = - 1$</p><p>Hence,${t}_{n} = {\left(\dfrac{{n}^{2}}{2} + \dfrac{3n}{2} - 1\right)}^{2}$</p><p>$\alpha = \mathop{∑}_{n = 1}^{10}{\left(\dfrac{{n}^{2}}{2} + \dfrac{3n}{2} - 1\right)}^{2}$</p><p>$⇒ 4\alpha = \mathop{∑}_{n = 1}^{10}{\left({n}^{2} + 3n - 2\right)}^{2} and $\beta = \mathop{∑}_{n = 1}^{10}{n}^{4}$

$⇒ 4\alpha - \beta = \mathop{∑}_{n = 1}^{10}\left(6{n}^{3} + 5{n}^{2} - 12n + 4\right)$

$⇒ 4\alpha - \beta = \left(6\mathop{∑}_{n = 1}^{10}{n}^{3} + 5\mathop{∑}_{n = 1}^{10}{n}^{2} - 12\mathop{∑}_{n = 1}^{10}n + 4\mathop{∑}_{n = 1}^{10}1\right)$

$⇒ 4\alpha - \beta = \left(6{\left(\dfrac{10\left(10 + 1\right)}{2}\right)}^{2} + 5\left(\dfrac{10 \cdot 11 \cdot 21}{6}\right) - 12\left(\dfrac{10 \cdot 11}{2}\right) + 4 \cdot 10\right)$

$⇒ 4\alpha - \beta = \left(6{\left(55\right)}^{2} + 5 \cdot 385 - 12 \cdot 55 + 40\right)$

$⇒ 4\alpha - \beta = 55 ( 353 ) + 40$

Hence, on comparing with $55k + 40$ we get, $k = 353$