Let =1^2+4^2+8^2+13^2+19^2+26^2+ upto 10 terms and = _n=1^10n^4.

Question

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 $4\alpha -\beta =55k+40$, then $k$ is equal to \_\_\_\_\_\_\_\_\_\_.

Answer: 353

Accepted Answer

$\begin{gathered} \alpha ={1}^{2}+{4}^{2}+{8}^{2}\ldots . \ {t}_{n}={a}^{2}+bn+c \end{gathered}$

$\begin{gathered} 1=a+b+c \ 4=4a+2b+c \ 8=9a+3b+c \end{gathered}$

On solving we get, $a=\frac{1}{2},b=\frac{3}{2},c=-1$

$\begin{gathered} \alpha =\sum _{n=1}^{10}{(\frac{{n}^{2}}{2}+\frac{3n}{2}-1)}^{2} \ 4\alpha =\sum _{n=1}^{10}{({n}^{2}+3n-2)}^{2},\beta =\sum _{n=1}^{10}{n}^{4} \ 4\alpha -\beta =\sum _{n=1}^{10}(6{n}^{3}+5{n}^{2}-12n+4)=55(353)+40 \end{gathered}$

Solutions · 1