Let sum_{k=1}^{n} a_{k} = alpha n^{2} + beta | vedclass

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(C) Given the sum of $n$ terms $S_n = \alpha n^2 + \beta n$.We know that $a_n = S_n - S_{n-1}$.$a_n = (\alpha n^2 + \beta n) - (\alpha(n-1)^2 + \beta(

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$5$

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(C) Given the sum of $n$ terms $S_n = \alpha n^2 + \beta n$.We know that $a_n = S_n - S_{n-1}$.$a_n = (\alpha n^2 + \beta n) - (\alpha(n-1)^2 + \beta(n-1))$$a_n = \alpha n^2 + \beta n - (\alpha(n^2 - 2n + 1) + \beta n - \beta)$$a_n = 2\alpha n - \alpha + \beta$.Given $a_{10} = 59$, so $2\alpha(10) - \alpha + \beta = 59 \Rightarrow 19\alpha + \beta = 59$ (Equation $1$).Given $a_6 = 7a_1$, so $2\alpha(6) - \alpha + \beta = 7(2\alpha(1) - \alpha + \beta)$.$11\alpha + \beta = 7(\alpha + \beta) \Rightarrow 11\alpha + \beta = 7\alpha + 7\beta$.$4\alpha = 6\beta$ $\Rightarrow 2\alpha = 3\beta$ $\Rightarrow \beta = \frac{2}{3}\alpha$ (Equation $2$).Substitute Equation $2$ into Equation $1$:$19\alpha + \frac{2}{3}\alpha = 59$ $\Rightarrow \frac{57\alpha + 2\alpha}{3} = 59$ $\Rightarrow \frac{59\alpha}{3} = 59$ $\Rightarrow \alpha = 3$.Then $\beta = \frac{2}{3}(3) = 2$.Therefore, $\alpha + \beta = 3 + 2 = 5$.

Let $\sum_{k=1}^{n} a_{k} = \alpha n^{2} + \beta n$. If $a_{10} = 59$ and $a_{6} = 7a_{1}$, then $\alpha + \beta$ is equal to:

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