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

Correct Answer: 353

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:

\[ k = 353 \]