Let $\alpha=1^2+4^2+8^2+13^2+19^2+26^2+\ldots$ upto 10 terms and $\beta=\sum_\limits{n=1}^{10} n^4$. If $4 \alpha-\beta=55 k+40$, then $\mathrm{k}$ is equal to __________.
Answer (integer)
353
Solution
<p>$$\begin{gathered}
\alpha=1^2+4^2+8^2 \ldots . \\
t_n=a^2+b n+c
\end{gathered}$$</p>
<p>$$\begin{aligned}
& 1=a+b+c \\
& 4=4 a+2 b+c \\
& 8=9 a+3 b+c
\end{aligned}$$</p>
<p>On solving we get, $\mathrm{a}=\frac{1}{2}, \mathrm{~b}=\frac{3}{2}, \mathrm{c}=-1$</p>
<p>$$\begin{aligned}
& \alpha=\sum_{n=1}^{10}\left(\frac{n^2}{2}+\frac{3 n}{2}-1\right)^2 \\
& 4 \alpha=\sum_{n=1}^{10}\left(n^2+3 n-2\right)^2, \beta=\sum_{n=1}^{10} n^4 \\
& 4 \alpha-\beta=\sum_{n=1}^{10}\left(6 n^3+5 n^2-12 n+4\right)=55(353)+40
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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