Question

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false.
1^2 + 4^2 + 7^2 + ... + (3n-2)^2 = (n(6n^2 - 3n - 1))/2

Answer 1

First check if it's true for \(n=1\) (the base case).
\[\frac{1(6\times1^2-3\times1-1)}{2}=\frac{6-3-1}{2}\stackrel{\checkmark}=1\]
Assume the statement is true for \(n=k\), i.e.
\[1^2+4^2+\cdots+(3k-2)^2=\frac{k(6k^2-3k-1)}{2}\]
You want to use this hypothesis to establish that
\[1^2+4^2+\cdots+(3k-2)^2+(3(k+1)-2)^2=\frac{(k+1)(6(k+1)^2-3(k+1)-1)}{2}\]
From the hypothesis, you know that the first \(k\) terms on the left can be simplified:
\[\frac{k(6k^2-3k-1)}{2}+(3(k+1)-2)^2=\frac{(k+1)(6(k+1)^2-3(k+1)-1)}{2}\]
Some rewriting of the right side:
\[\frac{k(6k^2-3k-1)}{2}+(3(k+1)-2)^2=\frac{(k+1)(6k^2+9k+2)}{2}\]
Now it's a matter of establishing that this here is an identity.