Mathematics - Sequence & Series Question with Solution | TestHub

Given,

$\alpha =1^2+4^2+8^2\dots$

Now, let $\sqrt{t_n}=a{n}^2+bn+c$

$1=a+ b+c ....\left(i\right)$

$4=4a+2b+c ....\left(ii\right)$

$8=9a+3b+c ....\left(iii\right)$

Now, solving equation $\left(i\right), \left(ii\right) \& \left(iii\right)$ we get,

$a=\frac{1}{2}, b=\frac{3}{2}, c=-1$

Hence, $t_n={\left(\frac{n^2}{2}+\frac{3n}{2}-1\right)}^2$

$\alpha =\sum _{n=1}^{10}{\left(\frac{n^2}{2}+\frac{3n}{2}-1\right)}^2$

$\Rightarrow 4\alpha =\sum _{n=1}^{10}{\left(n^2+3n-2\right)}^2$ and $\beta =\sum _{n=1}^{10}{n}^4$

$\Rightarrow 4\alpha -\beta =\sum _{n=1}^{10}\left(6n^3+5n^2-12n+4\right)$

$\Rightarrow 4\alpha -\beta =\left(6\sum _{n=1}^{10}{n}^3+5\sum _{n=1}^{10}{n}^2-12\sum _{n=1}^{10}n+4\sum _{n=1}^{10}1\right)$

$\Rightarrow 4\alpha -\beta =\left(6{\left(\frac{10\left(10+1\right)}{2}\right)}^2+5\left(\frac{10·11·21}{6}\right)-12\left(\frac{10·11}{2}\right)+4·10\right)$

$\Rightarrow 4\alpha -\beta =\left(6{\left(55\right)}^2+5·385-12·55+40\right)$

$\Rightarrow 4\alpha -\beta =55\left(353\right)+40$

Hence, on comparing with $55k+40$ we get, $k=353$