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 _______.

Let the terms inside the squares for $$\alpha$$ be $$t_n$$. The sequence is $$1, 4, 8, 13, 19, 26, \dots$$

The first differences are $$3, 4, 5, 6, 7, \dots$$ which form an arithmetic progression. This means $$t_n$$ is a quadratic expression.

Let $$t_n = an^2 + bn + c$$. Using the first three terms:

• $$t_1 = a + b + c = 1$$
• $$t_2 = 4a + 2b + c = 4$$
• $$t_3 = 9a + 3b + c = 8$$

Solving these equations gives $$a = \frac{1}{2}$$, $$b = \frac{3}{2}$$, and $$c = -1$$.

The general term is:

$$t_n = \frac{n^2 + 3n - 2}{2}$$

The $$n$$th term of $$\alpha$$ is $$t_n^2$$:

$$t_n^2 = \left( \frac{n^2 + 3n - 2}{2} \right)^2 = \frac{n^4 + 6n^3 + 5n^2 - 12n + 4}{4}$$

We are given $$\alpha = \sum_{n=1}^{10} t_n^2$$ and $$\beta = \sum_{n=1}^{10} n^4$$. We need to evaluate $$4\alpha - \beta$$:

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

Subtracting $$\beta$$:

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

Apply standard summation formulas for $$n = 10$$:

• $$\sum n^3 = \left( \frac{10 \times 11}{2} \right)^2 = 3025$$
• $$\sum n^2 = \frac{10 \times 11 \times 21}{6} = 385$$
• $$\sum n = \frac{10 \times 11}{2} = 55$$
• $$\sum 4 = 4 \times 10 = 40$$

Substitute these values back into the expression:

$$4\alpha - \beta = 6(3025) + 5(385) - 12(55) + 40$$
$$4\alpha - \beta = 18150 + 1925 - 660 + 40 = 19455$$

We are given that $$4\alpha - \beta = 55k + 40$$. Equating the two:

$$55k + 40 = 19455$$ $$55k = 19415$$ $$k = 353$$