Question

Please help me to solve this mathematical induction problem, 2^2+5^2+8^2+….+(3n-1)^2=1n/2(6n^2+3n-1)

Answer 1

Have you gone through the steps (e.g. assume true for k...)

Answer 2

Proof: For each positive integer \(n\ge1\) , let \(S(n)\) be the statement
\[2^2+5^2+8^2+...+(3n-1)^2=\frac{n}{2}(6n^2+3n-1)\]
Basis step: \(S(1)\) is the statement \(2^2=\frac{1}{2}(6+3-1)=4\). Thus \(S(1)\) is true.

Inductive step: We suppose that \(S(k)\) is true and prove that \(S(k+1)\) is true.

Thus, we assume that
\[2^2+5^2+8^2+...+(3k-1)^2=\frac{k}{2}(6k^2+3k-1)\]
and prove that
\[2^2+5^2+8^2+...+(3k+1)^2=\frac{k+1}{2}(6(k+1)^2+3(k+1)-1)\]

Answer 3

lol, I dont see how any of this helps

Answer 4

but it looks cool

Answer 5

me too i dont see...lol :)

Answer 6

ok @nroprstar

u need to show that\[\frac{k}{2}(6k^2+3k-1)+(3k+1)^2=\frac{k+1}{2}(6(k+1)^2+3(k+1)-1)\]why and how??? this is ur work !! :)

Answer 7

oops thats\[\frac{k}{2}(6k^2+3k-1)+(3k+2)^2=\frac{k+1}{2}(6(k+1)^2+3(k+1)-1)\]