Let alpha = 1^2 + 4^2 + 8^2 + 13^2 + 19^2 + 2 | vedclass

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(B) The $n$-th term of the sequence $1, 4, 8, 13, 19, 26, \ldots$ is given by $a_n = \frac{n^2+3n-2}{2}$.Thus,$\alpha = \sum_{n=1}^{10} \left(\frac{n^

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$353$

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(B) The $n$-th term of the sequence $1, 4, 8, 13, 19, 26, \ldots$ is given by $a_n = \frac{n^2+3n-2}{2}$.Thus,$\alpha = \sum_{n=1}^{10} \left(\frac{n^2+3n-2}{2}\right)^2 = \frac{1}{4} \sum_{n=1}^{10} (n^2+3n-2)^2$.Then $4\alpha = \sum_{n=1}^{10} (n^4 + 9n^2 + 4 + 6n^3 - 4n^2 - 12n) = \sum_{n=1}^{10} (n^4 + 6n^3 + 5n^2 - 12n + 4)$.Given $\beta = \sum_{n=1}^{10} n^4$,we have $4\alpha - \beta = \sum_{n=1}^{10} (6n^3 + 5n^2 - 12n + 4)$.Using standard summation formulas:$\sum_{n=1}^{10} n^3 = (55)^2 = 3025$,$\sum_{n=1}^{10} n^2 = \frac{10(11)(21)}{6} = 385$,$\sum_{n=1}^{10} n = 55$.$4\alpha - \beta = 6(3025) + 5(385) - 12(55) + 4(10) = 18150 + 1925 - 660 + 40 = 19455$.We are given $4\alpha - \beta = 55k + 40$,so $55k = 19455 - 40 = 19415$.$k = \frac{19415}{55} = 353$.

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 . . . . . . .

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