Question

2.12 + 42 + 72 + ... + (3n - 2)2 = quantity n times quantity six n squared minus three n minus one all divided by two

Answer 1

@satellite73

Answer 2

I have to state whether the statement is true or false for all positive integers n but Idk how to do this problem, it doesn't make sense

Answer 3

@wio

Answer 4

This?\[
12+42+72+\ldots +(3n-2)^2=n\left(\frac{6n^2-3n-1}{2}\right)
\]

Answer 5

yes:)

Answer 6

But \((3(1)-2)^2 = (3-2)^2 = 1^2=1\)

Answer 7

And \[
(3(2)-2)^2 = (6-2)^2 = 4^2=16
\]

Answer 8

\[
(3(3)-2)^2 = (9-2)^2 = 7^2 = 49
\]

Answer 9

good point
none of these numbers are squares
is there a typo somewhere?

Answer 10

OH SORRY its supposed to be 1^2, 4^2, 7^2, ...

Answer 11

Okay, so the first step is to let \(n=1\) and then simplify: \[
n\left(\frac{6n^2-3n-1}{2}\right)
\]

Answer 12

\[
1\left(\frac{6(1)^2-3(1)-1}{2}\right) = \frac{6-3-1}{2} = \frac 22 = 1
\]So it works for \(n=1\).

Answer 13

ahh ok I wrote my numbers wrong when I wrote the problem. so the statement would be true, correct?

Answer 14

hmm is this supposed to be a proof by induction, or do you just need to check that
\[\sum_{k=1}^n(3k-2)^2=n\left(\frac{6n^2-3n-1}{2}\right)\]

Answer 15

\[
2\left(\frac{6(2)^2-3(2)-1}{2}\right) = 6(4)-3(2)-1 = 24-6-1 = 17
\]So it doesn't seem to work for \(n=2\).
Unless I made a mistake.

Answer 16

Wait a moment

Answer 17

\[1 +16 = 17\] So actually it does work for \(n=2\)

Answer 18

so would the statement be true?