Mathematics - Sequence & Series Question with Solution | TestHub

Given,

$S=1^2-2·3^2+3·5^2............+15·{29}^2$

Now rewriting the above expression we get,

$S=\underset{S_1}{\underbrace{1^2+2·3^2+3·5^2............+15·{29}^2}}-2\underset{S_2}{\underbrace{\left[2·3^2+4·7^2............+14·{27}^2\right]}}$

Now solving $S_1=1^2+2·3^2+3·5^2............+15·{29}^2$

$\Rightarrow S_1=\sum _{r=1}^{15}r·{\left(2r-1\right)}^2$

$\Rightarrow S_1=\sum _{r=1}^{15}r·\left(4r^2+1-4r\right)$

$\Rightarrow S_1=\sum _{r=1}^{15}\left(4r^3+r-4r^2\right)$

$\Rightarrow S_1=\left(4\sum _{r=1}^{15}{r}^3+\sum _{r=1}^{15}r-4\sum _{r=1}^{15}{r}^2\right)$
$\Rightarrow S_1=\left(4{\left(\frac{15\times 16}{2}\right)}^2+\frac{15\times 16}{2}-4\frac{15\times 16\times 31}{6}\right)$

$\Rightarrow S_1=\left(4{\left(120\right)}^2+120-4960\right)$

$\Rightarrow S_1=52760$

Now solving $S_2=2·3^2+4·7^2............+14·{27}^2$

$\Rightarrow S_2=\sum _{r=1}^7\left(2r\right){\left(4r-1\right)}^2$

$\Rightarrow S_2=\sum _{r=1}^7\left(2r\right)\left(16r^2+1-8r\right)$

$\Rightarrow S_2=32\sum _{r=1}^7{r}^3+2\sum _{r=1}^{7}r-16\sum _{r=1}^7{r}^2$

$\Rightarrow S_2=32{\left(\frac{7\times 8}{2}\right)}^2+2\left(\frac{7\times 8}{2}\right)-16\left(\frac{7\times 8\times 15}{6}\right)$

$\Rightarrow S_2=32\times 784+56-16\times 140=22904$

Hence, $S=S_1-2S_2=52760-2\times 22904=6952$