For this kind of problem I would use a proving method known as induction. An inductive proof has two steps: a *basis step* and an *inductive step*.

The basis step is where you prove the statement is true for the smallest possible value.

For the inductive step, you prove that *if* the statement is true for some arbitrary value, then it is also true for the very next value.

With these both combined, you will have proven it is true for all values.

For your example problem, we will start with the basis step and show that it is true for the smallest value of n, where n = 1:

Σ^(n)\_k=1 (4k - 3) = 2n^(2) \- n

Σ^(1)\_k=1 (4k - 3) = 2(1)^(2) \- 1

4(1) - 3 = 2(1) - 1

4 - 3 = 2 - 1

1 = 1

Done.

Next, we will say that, if it is true for some value n = p, then it is also true for n = p + 1. And here's how I would personally go about that:

Given:

Σ^(p)\_k=1 (4k - 3) = 2p^(2) \- p

Then:

Σ^(p+1)\_k=1 (4k - 3)

= (Σ^(p)\_k=1 (4k - 3)) + (4(p+1) - 3)

= 2p^(2) \- p + (4(p+1) - 3) \[Substitution with our Given\]

= 2p^(2) \- p + (4p + 4 - 3)

= 2p^(2) \- p + 4p + 1

= 2p^(2) \- p + 4p + 1 + 1 - 1

= 2p^(2) \- p + 4p + 2 - 1

= 2p^(2) \+ 4p + 2 - p - 1

= 2(p^(2) \+ 2p + 1) - (p + 1)

= 2(p + 1)^(2) \- (p + 1)

And there you have it, starting with:

Σ^(p)\_k=1 (4k - 3) = 2p^(2) \- p

We showed that:

Σ^(p+1)\_k=1 (4k - 3) = 2(p + 1)^(2) \- (p + 1)