Let $\alpha=1^2+4^2+8^2+13^2+19^2+26^2+\cdots$ up to 10 terms and $\beta=\displaystyle\sum_{n=1}^{10}n^4$. If $4\alpha-\beta=55k+40$, then $k$ is equal to
Step-by-Step Solution
Key Concept: Find the general term $t_n$ of $1,4,8,13,19,26,\ldots$ (differences: 3,4,5,6,7,...). $t_n=\frac{n^2+3n-2}{2}$. Compute $4\alpha=4\sum_{n=1}^{10}t_n^2$ and $\beta=\sum n^4$, then $4\alpha-\beta=55k+40$.
$4\alpha-\beta=55\times353+40$. $k=353$.
Correct Answer: 353