Show by the Principle of Mathematical induction that the sum

Let $\mathrm{P}(n): S_{n}=1^{2}+2 \times 2^{2}+3^{2}+2 \times 4^{2}+5^{2}+\ldots=\left\{\begin{array}{l}\frac{n(n+1)^{2}}{2}, \text { when } n \text { is even } \\ \frac{n^{2}(n+1)}{2}, \text { when } n \text { is odd }\end{array}\right.$

Step I: For $n=1$ i.e. $\mathrm{P}(1)$ :

$\mathrm{LHS}=\mathrm{S}_{1}=1^{2}=1$

$\mathrm{RHS}=\mathrm{S}_{1}=\frac{1^{2}(1+1)}{2}=1$

As, LHS $=$ RHS

So, it is true for $n=1$.

Step II : For $n=k$,

Let $\mathrm{P}(k): S_{k}=1^{2}+2 \times 2^{2}+3^{2}+2 \times 4^{2}+5^{2}+\ldots=\left\{\begin{array}{l}\frac{k(k+1)^{2}}{2}, \text { when } k \text { is even }, \text { be true for } \\ \frac{k^{2}(k+1)}{2}, \text { when } k \text { is odd }\end{array}\right.$

some natural numbers.

Step III : For $n=k+1$,

Case 1: When $k$ is odd, then $(k+1)$ is even.

$\mathrm{P}(k+1):$

$\mathrm{LHS}=S_{k+1}=1^{2}+2 \times 2^{2}+3^{2}+2 \times 4^{2}+5^{2}+\ldots+k^{2}+2 \times(k+1)^{2}$

$=\frac{k^{2}(k+1)}{2}+2 \times(k+1)^{2} \quad$ (Using step II)

$=\frac{k^{2}(k+1)+4(k+1)^{2}}{2}$

$=\frac{(k+1)\left(k^{2}+4 k+4\right)}{2}$

$=\frac{(k+1)(k+2)^{2}}{2}$

$\mathrm{RHS}=\frac{(k+1)(k+1+1)^{2}}{2}$

$=\frac{(k+1)(k+2)^{2}}{2}$

As, LHS = RHS

So, it is true for $n=k+1$ when $k$ is odd.

Case 2: When $k$ is even, then $(k+1)$ is odd.

$\mathrm{P}(k+1):$

$\mathrm{RHS}=S_{k+1}=1^{2}+2 \times 2^{2}+3^{2}+2 \times 4^{2}+5^{2}+\ldots+2 \times k^{2}+(k+1)^{2}$

$=\frac{k(k+1)^{2}}{2}+(k+1)^{2} \quad$ (Using step II)

$=\frac{k(k+1)^{2}+2(k+1)^{2}}{2}$

$=\frac{(k+1)^{2}(k+2)}{2}$

$=\frac{(k+1)^{2}(k+2)}{2}$

$\mathrm{RHS}=\frac{(k+1)^{2}(k+1+1)}{2}$

$=\frac{(k+1)^{2}(k+2)}{2}$

As, LHS $=$ RHS

So, it is true for $n=k+1$ when $k$ is even.

Hence, $\mathrm{P}(n)$ is true for all natural numbers.