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

Sequence Identification for \( \alpha \): The sequence \( \alpha = \{1, 4, 8, 13, 19, 26, \dots\} \) exhibits constant second differences, indicating it is a quadratic sequence. The general term is given by \( T_n = an^2 + bn + c \). Using the initial terms \( T_1 = 1, T_2 = 4, T_3 = 8 \), the following system of equations is established:

\[ a + b + c = 1 \]

\[ 4a + 2b + c = 4 \]

\[ 9a + 3b + c = 8 \]

Solving this system yields \( a = \frac{1}{2}, b = \frac{3}{2}, c = -1 \). The general term for \( \alpha \) is \( T_n = \frac{1}{2}n^2 + \frac{3}{2}n - 1 \). Consequently, \( \alpha = \sum_{n=1}^{10} \left( \frac{1}{2}n^2 + \frac{3}{2}n - 1 \right)^2 \).

Expression for \( 4\alpha \): Simplifying \( 4\alpha \) results in \( 4\alpha = \sum_{n=1}^{10} (n^2 + 3n - 2)^2 \).

Calculation of \( \beta \): \( \beta = \sum_{n=1}^{10} n^4 \) is computed directly.

Determination of \( k \): Substituting into the equation \( 4\alpha - \beta = 55k + 40 \), and solving for \( k \), we obtain \( k = 353 \).