## Precalculus (10th Edition)

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 $\fbox{true}$.