Sequences And Series Question 13
Question 13 - 2024 (30 Jan Shift 1)
Let $\alpha=1^{2}+4^{2}+8^{2}+13^{2}+19^{2}+26^{2}+\ldots$ upto 10 terms and $\beta=\sum_{n=1}^{10} n^{4}$. If $4 \alpha-\beta=55 k+40$, then $\mathrm{k}$ is equal to
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Answer (353)
Solution
$\alpha=1^{2}+4^{2}+8^{2} \ldots$
$t_{n}=a^{2}+b n+c$
$1=a+b+c$
$4=4 a+2 b+c$
$8=9 a+3 b+c$
On solving we get, $\mathrm{a}=\frac{1}{2}, \mathrm{~b}=\frac{3}{2}, \mathrm{c}=-1$
$\alpha=\sum_{\mathrm{n}=1}^{10}\left(\frac{\mathrm{n}^{2}}{2}+\frac{3 \mathrm{n}}{2}-1\right)^{2}$
$4 \alpha=\sum_{n=1}^{10}\left(n^{2}+3 n-2\right)^{2}, \beta=\sum_{n=1}^{10} n^{4}$
$4 \alpha-\beta=\sum_{n=1}^{10}\left(6 n^{3}+5 n^{2}-12 n+4\right)=55(353)+40$
