Answer
See proof
Work Step by Step
We have to prove the statement:
$1^2+4^2+7^2+.....+(3n-2)^2=\dfrac{1}{2}n(6n^2-3n-1)$
Step 1: Prove that the statement is true for $n=1$:
The left side is:
$(3(1)-2)^2=1^2=1$
The right side is:
$\dfrac{1}{2}\cdot 1\cdot(6(1^2)-3(1)-1)=\dfrac{1}{2}\cdot 2=1$
As $1=1$, the statement is true for $n=1$.
Step 2:
We have to prove:
$1^2+4^2+7^2+.....+(3k-2)^2+(3(k+1)-2)^2=\dfrac{1}{2}(k+1)(6(k+1)^2-3(k+1)-1)$
$1^2+4^2+7^2+.....+(3k-2)^2+(3(k+1)-2)^2=\dfrac{1}{2}(k+1)(6(k^2+2k+1)-3k-3-1)$
$1^2+4^2+7^2+.....+(3k-2)^2+(3(k+1)-2)^2=\dfrac{1}{2}(k+1)(6k^2+12k+6-3k-4)$
$1^2+4^2+7^2+.....+(3k-2)^2+(3(k+1)-2)^2=\dfrac{1}{2}(k+1)(6k^2+9k+2)$
Because the statement is true for $k$ we have:
$1^2+4^2+7^2+.....+(3k-2)^2=\dfrac{1}{2}k(6k^2-3k-1)$
Add $(3(k+1)-2)^2$ on each side:
$1^2+4^2+7^2+.....+(3k-2)^2+(3(k+1)-2)^2=\dfrac{1}{2}k(6k^2-3k-1)+(3(k+1)-2)^2$
$1^2+4^2+7^2+.....+(3k-2)^2+(3(k+1)-2)^2=\dfrac{1}{2}k(6k^2-3k-1)+(3k+1)^2$
$1^2+4^2+7^2+.....+(3k-2)^2+(3(k+1)-2)^2=\dfrac{1}{2}(6k^3-3k^2-k+2(9k^2+6k+1)$
$1^2+4^2+7^2+.....+(3k-2)^2+(3(k+1)-2)^2=\dfrac{1}{2}(6k^3-3k^2-k+18k^2+12k+2)$
$1^2+4^2+7^2+.....+(3k-2)^2+(3(k+1)-2)^2=\dfrac{1}{2}(6k^3+15k^2+11k+2)$
$1^2+4^2+7^2+.....+(3k-2)^2+(3(k+1)-2)^2=\dfrac{1}{2}(6k^3+6k^2+9k^2+9k+2k+2)$
$1^2+4^2+7^2+.....+(3k-2)^2+(3(k+1)-2)^2=\dfrac{1}{2}(6k^2(k+1)+9k(k+1)+2(k+1)$
$1^2+4^2+7^2+.....+(3k-2)^2+(3(k+1)-2)^2=\dfrac{1}{2}(k+1)(6k^2+9k+2)$
We have proved that the statement is also true for $k+1$.
So, the statement is $\fbox{true}$.