Let $$\alpha = 1^2 + 4^2 + 8^2 + 13^2 + 19^2 + 26^2 + \ldots$$ upto $$10$$ terms and $$\beta = \sum_{n=1}^{10} n^4$$. If $$4\alpha - \beta = 55k + 40$$, then $$k$$ is equal to _______.

The given sequence inside $$\alpha$$ is $$1,4,8,13,19,26,\ldots$$.
The first term is $$a_1 = 1$$ and the successive differences are

$$a_2-a_1 = 3,\; a_3-a_2 = 4,\; a_4-a_3 = 5,\ldots$$

The $$n^{\text{th}}$$ difference is therefore $$d_n = n+1$$ for $$n \ge 2$$. Hence

$$a_n = a_{n-1} + (n+1),\qquad a_1 = 1$$

Summing the differences from $$k = 2$$ to $$k = n$$ gives

$$a_n = 1 + \sum_{k=2}^{n}(k+1)$$

Break the right-hand sum into two parts:

$$\sum_{k=2}^{n}k = \frac{n(n+1)}{2}-1,\qquad \sum_{k=2}^{n}1 = n-1$$

Therefore

$$a_n = 1 + \bigl(\tfrac{n(n+1)}{2}-1\bigr) + (n-1)$$

$$\phantom{a_n}= \frac{n(n+1)}{2} + n - 1 = \frac{n(n+3)}{2} - 1$$

So

$$\alpha = \sum_{n=1}^{10} a_n^{\,2} = \sum_{n=1}^{10}\left(\frac{n(n+3)}{2}-1\right)^{2}$$

Put $$b_n = \dfrac{n(n+3)}{2}-1 = \dfrac{n^{2}+3n-2}{2}$$.
Then

$$b_n^{2} = \frac{(n^{2}+3n-2)^{2}}{4}$$

Expand the square:

$$(n^{2}+3n-2)^{2} = n^{4} + 6n^{3} + 5n^{2} - 12n + 4$$

Hence

$$\alpha = \frac14 \sum_{n=1}^{10} \bigl(n^{4} + 6n^{3} + 5n^{2} - 12n + 4\bigr)$$

Multiply by $$4$$:

$$4\alpha = \sum_{n=1}^{10} \bigl(n^{4} + 6n^{3} + 5n^{2} - 12n + 4\bigr)$$

Given $$\beta = \displaystyle\sum_{n=1}^{10} n^{4}$$, subtracting $$\beta$$ eliminates the $$n^{4}$$ terms:

$$4\alpha - \beta = \sum_{n=1}^{10} \bigl(6n^{3} + 5n^{2} - 12n + 4\bigr)$$

Now evaluate each standard sum up to $$10$$:

$$\sum_{n=1}^{10} n = 55$$

$$\sum_{n=1}^{10} n^{2} = \frac{10\cdot11\cdot21}{6} = 385$$

$$\sum_{n=1}^{10} n^{3} = \left(\frac{10\cdot11}{2}\right)^{2} = 3025$$

$$\sum_{n=1}^{10} 1 = 10$$

Substitute into the expression for $$4\alpha - \beta$$:

$$4\alpha - \beta = 6(3025) + 5(385) - 12(55) + 4(10)$$

$$\phantom{4\alpha - \beta}= 18150 + 1925 - 660 + 40$$

$$\phantom{4\alpha - \beta}= 19455$$

The problem states $$4\alpha - \beta = 55k + 40$$. Equate:

$$55k + 40 = 19455$$

$$55k = 19415$$

$$k = \frac{19415}{55} = 353$$

Therefore, $$k = 353$$.