Let = 1^2+4^2+8^2+13^2+19^2+26^2+ up to 10 terms, and let be a related quantity defined in…

Question

Let $\alpha = 1^2+4^2+8^2+13^2+19^2+26^2+\cdots$ up to 10 terms, and let $\beta$ be a related quantity defined in the source problem. If $4\alpha-\beta = 55k+40$, then $k$ is equal to \_\_\_\_\_\_\_\_.

Answer: 353

Accepted Answer

The base sequence $1,4,8,13,19,26,\ldots$ has second-order constant differences ($3,4,5,6,\ldots$), giving $T_n = 	frac12n^2+	frac32n-1$, so $\alpha=\sum_{n=1}^{10}T_n^2 = 11197$; the stated final answer is $k=353$.

Solutions · 1