Question

Using mathematical induction prove whether or not the following statement is true for all positive integers n, or show why it is false.
2+6+10+ ... +4n-2=2n^2

Answer 1

Base case: For $n=1$, the left side is 2 and the right is $2\cdot1^2=2$, so the base case holds.

Induction hypothesis: Assume the statement is true for $n=k$, that is

$2+6+10+\cdots+4k-2=2k^2$

We want to show that this implies truth for $n=k+1$, that

$2+6+10+\cdots+4k-2+4(k+1)-2=2(k+1)^2$

The first $k$ terms on the left reduce according to the assumption above, and we can simplify the $k+1$-th term a bit:

$\underbrace{2+6+10+\cdots+4k-2}_{2k^2}+4k+2$

$2k^2+4k+2=2(k^2+2k+1)=2(k+1)^2$

so the statement is true for all $n\in\mathbb N$.