Question

Prove by induction, 2? Thanks :)

Answer 1

Let us assume that the statement is true for some `n`, i.e, for some `n` the relation

`sum_{i=1}^n (4i-2) = 2n^2`

is true. Then

`sum_{i=1}^{n+1} (4i-2) =[sum_{i=1}^{n} (4i-2)]+4(n+1)-2`
`qquad = 2n^2+4n+2 = 2(n^2+2n+1) = 2(n+1)^2`

Thus, if the statement is true for `n`, it is true for `n+1`.

Now, the statement is obviously true for `n=1`. Hence, by mathematical induction, it is true for all natural numbers `n`.