Question

prove using mathematical induction
12 + 42 + 72 + ... + (3n - 2)2 = (n(〖6n〗^2-3n-1))/2

Answer 1

What?

Answer 2

start with the first term, then your IH will be that if that holds for n, it must hold for n+1 also (f(n)=g(n) => f(n+1)=g(n+1))

Answer 3

Here's what I have so far:
\[s _{1}=\frac{ n(6n ^{2}-3n-1) }{2}\]
\[=\frac{ 1(6(1^{2})^{2}-3(1)-1) }{ 2 }\]
\[=\frac{ 1(2) }{ 2 }\]
\[S _{1}=1\]

Answer 4

@angela101

Answer 5

@courtney_celiene1236
Good, that's a good first step.

By the way, we at OpenStudy here write
\(1^2\) as 1^2. It's ugly, but more understandable than 12.

If we denote:
\(f(x)=1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 \)
and
\(\large g(x)=\frac{n(6n^2-3n-1)}{2}\)
then the next step you need to do is to show that
"given f(n)=g(n), then g(n+1)=f(n+1)" ...............(1)
this is the same as proving
g(n+1)-g(n)=f(n+1)-f(n) .....................(2)

From the definition of f(n), we know that
\(f(n+1)=f(n)+(3(n+1)-2)^2\)
which means that
\(f(n+1)-f(n)=(3(n+1)-2)^2\)
that reduces (2) to proving
\(g(n+1)-g(n)=(3(n+1)-2)^2\) ................(3)
So expanding (3) is all you need to complete the proof.