Let _k=1^n a_k = n^2 + n for all n.

Question

Let $\displaystyle\sum_{k=1}^{n} a_k = \alpha n^2 + \beta n$ for all $n$. If $a_{10}=59$ and $a_6 = 7a_1$, then $\alpha+\beta$ is equal to:

Accepted Answer

Since $S_n=\alpha n^2+\beta n$ has no constant term, $\{a_n\}$ is an A.P., and
$$a_n = S_n - S_{n-1} = \beta-\alpha+2\alpha n.$$

$a_{10}=59 \Rightarrow \beta-\alpha+20\alpha = 59 \Rightarrow 19\alpha+\beta=59.$

$a_6 = 7a_1 \Rightarrow \beta-\alpha+12\alpha = 7(\beta-\alpha+2\alpha) \Rightarrow 11\alpha+\beta = 7\alpha+7\beta \Rightarrow 4\alpha = 6\beta \Rightarrow \alpha = \tfrac{3\beta}{2}.$

Substituting: $19\left(\tfrac{3\beta}{2}\right)+\beta = 59 \Rightarrow \tfrac{57\beta+2\beta}{2}=59 \Rightarrow 59\beta = 118 \Rightarrow \beta=2$, so $\alpha = 3$.

$$\alpha+\beta = 3+2 = 5.$$

Solutions · 1