Question

Find the sum of the following series to $n$ terms :
$2^2+4^2+6^2+8^2+...$

Answer 1

Now given series $2^2+4^2+6^2+8^2+.........$

$\Rightarrow$ from above series

$\begin{array}{c}\Rightarrow 2^2\left(1^2+2^2+3^2+4^2+..........+n^2\right)\\ \Rightarrow 4\left(1^2+2^2+3^2+4^2+.........+n^2\right)\\ \Rightarrow 4p\left(n\right)\end{array}$

now $P\left(n\right)=1^2+2^2+3^2+4^2+..........+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$

from $n=1\Rightarrow P\left(1\right)=\frac{1\left(1+1\right)\left(2+1\right)}{6}=\frac{1\times 2\times 3}{6}=1$ that's true

now from $n=k\Rightarrow P\left(k\right)=\frac{k\left(k+1\right)\left(2k+1\right)}{6}\to \left(i\right)$

and again $n=k+1\Rightarrow P\left(k+1\right)=\frac{\left(k+1\right)\left(k+1+1\right)\left(2k+2+1\right)}{6}\to \left(ii\right)$

Now we proved $P(k+1)$ is true

then $\left(1^2+2^2+3^2+.......+k^2\right)+{\left(k+1\right)}^2$

using equation (i)

$\begin{array}{c}\Rightarrow \frac{k\left(k+1\right)\left(2k+1\right)}{6}+{\left(k+1\right)}^2=\frac{k\left(k+1\right)\left(2k+1\right)+6{\left(k+1\right)}^2}{6}\\ \Rightarrow \frac{k\left(k+1\right)\left(2k+1\right)+6{\left(k+1\right)}^2}{6}=\frac{\left(k+1\right)\left(2k^2+7k+6\right)}{6}=\\ \Rightarrow \frac{\left(k+1\right)\left(k+1+1\right)\left\{2\left(k+1\right)+1\right\}}{6}\to \left(iii\right)\end{array}$

from equation (ii) and (iii) $p(k+1)$ has true.