Question

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

Answer 1

Identify the Sequence for $\alpha$: The terms in $\alpha$ are 1, 4, 8, 13, 19, 26, ..., which represents a sequence with second differences that are constant. This indicates a quadratic sequence. Let the general term of this sequence be $T_n = an^2 + bn + c$. Using the terms:

$T_1 = 1, \quad T_2 = 4, \quad T_3 = 8$

Set up equations:

$a + b + c = 1$

$4a + 2b + c = 4$

$9a + 3b + c = 8$

Solving these, we get:

$a = \frac{1}{2}, \quad b = \frac{3}{2}, \quad c = -1$

General Term for $\alpha$: The n-th term of $\alpha$ is:

$T_n = \frac{1}{2}n^2 + \frac{3}{2}n - 1$

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

Expression for $4\alpha$: Expand and simplify $4\alpha = \sum_{n=1}^{10} (n^2 + 3n - 2)^2$.

Calculate $\beta$: $\beta = \sum_{n=1}^{10} n^4$, which can be computed directly.

Find $k$: Substitute into the expression:

$4\alpha - \beta = 55k + 40$

Solving for $k$, we find:

$k = 353$