2022-M2-03

For $n=1$,

$\begin{array}{cl}
& \text{LS} \\
= & \dsum_{k=1}^{2}(-1)^k k^2 \\
= & (-1)^1(1)^2+(-1)^2 (2)^2 \\
= & 3
\end{array}$

$\begin{array}{cl}
& \text{RS} \\
= & (1)[2(1)+1] \\
= & 3
\end{array}$

$\therefore \text{LS}=\text{RS}$.

Therefore, $P(1)$ is true.

Assume that $P(m)$ is true for some positive integer $m$.

i.e. $\dsum_{k=1}^{2m}(-1)^k k^2 =m(2m+1)$

For $n=m+1$,

$\begin{array}{cl}
& \text{LS} \\
= & \dsum_{k=1}^{2(m+1)} (-1)^k k^2 \\
= & \dsum_{k=1}^{2m}(-1)^k k^2 +(-1)^{2m+1}(2m+1)^2+(-1)^{2m+2}(2m+2)^2 \\
= & m(2m+1)-(2m+1)^2+(2m+2)^2 \\
= & (2m+1)(m-2m-1)+4(m+1)^2 \\
= & -(m+1)(2m+1)+4(m+1)^2 \\
= & (m+1)(-2m-1+4(m+1)) \\
= & (m+1)(2m+3) \\
= & \text{RS}
\end{array}$

Therefore, $P(m+1)$ is also true.

$\therefore$ by mathematical induction, $P(n)$ is true for all positive integers $n$.