Prove the following by using the principle of mathematical induction for all $n \in N:$
$4+8+12+\ldots+4n = 2n(n+1)$
$4+8+12+\ldots+4n = 2n(n+1)$
Let $P(n)$ be the statement $4+8+12+\ldots+4n = 2n(n+1)$.
Step $1$: For $n=1$,the left-hand side is $4(1) = 4$ and the right-hand side is $2(1)(1+1) = 2(2) = 4$.
Since $4=4$,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in N$,i.e.,$4+8+12+\ldots+4k = 2k(k+1)$.
Step $3$: We need to show $P(k+1)$ is true,i.e.,$4+8+12+\ldots+4k+4(k+1) = 2(k+1)(k+2)$.
Starting with the left-hand side:
$(4+8+12+\ldots+4k) + 4(k+1) = 2k(k+1) + 4(k+1)$
$= 2k^2 + 2k + 4k + 4$
$= 2k^2 + 6k + 4$
$= 2(k^2 + 3k + 2)$
$= 2(k+1)(k+2)$.
This matches the right-hand side for $n=k+1$.
Thus,by the principle of mathematical induction,$P(n)$ is true for all $n \in N$.
Step $1$: For $n=1$,the left-hand side is $4(1) = 4$ and the right-hand side is $2(1)(1+1) = 2(2) = 4$.
Since $4=4$,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in N$,i.e.,$4+8+12+\ldots+4k = 2k(k+1)$.
Step $3$: We need to show $P(k+1)$ is true,i.e.,$4+8+12+\ldots+4k+4(k+1) = 2(k+1)(k+2)$.
Starting with the left-hand side:
$(4+8+12+\ldots+4k) + 4(k+1) = 2k(k+1) + 4(k+1)$
$= 2k^2 + 2k + 4k + 4$
$= 2k^2 + 6k + 4$
$= 2(k^2 + 3k + 2)$
$= 2(k+1)(k+2)$.
This matches the right-hand side for $n=k+1$.
Thus,by the principle of mathematical induction,$P(n)$ is true for all $n \in N$.
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