Answer
please see step-by-step
Work Step by Step
Principle of Mathematical Induction$:$
If both following conditions are satisfied,
1. $P(1)$ is true.
2. For any $k\in \mathbb{N}$, [$P(k)$ is true] $\Rightarrow$ [$P(k+1)$ is true].
then
$P(n)$ is true for all $n\in \mathbb{N}$.
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Let $P(n): 2+4+6+\cdots+2n =n(n +1)$.
1: $P(1)$ : $2=1(1+1)$,
$P(1)$ is true.
2: Assume that $P(k)$ is true, that is,
$ 2+4+6+\cdots+2k=k(k+1)\qquad$(*)
(is $P(k+1)$ true?)
Add the next term of the LHS of (*) to both sides:
$2+4+6+\cdots+2k+2(k+1) = k(k+1)+2(k+1)$
factor out (k+1) on the RHS,
$2+4+6+\cdots+2k+2(k+1) = (k+1)(k+2)$
rewrite the RHS as it would be written in P(k+1)
$2+4+6+\cdots+2k+2(k+1) =(k+1)[(k+1)+1]$
Thus, $P(k)$ is true $\Rightarrow P(k+1)$ is true..
So by the Principle of Mathematical Induction,
$P(n)$ is true for all $n\in \mathbb{N}$.
