Question

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false.

4 x 6 + 5 x 7 + 6 x 8 +...+ 4n(4n+2) = 4(4n+1)(8n+7)/6

Answer 1

@Callisto

Answer 2

\[(4\cdot6)+(5\cdot7)+(6\cdot8)+\cdots+4n(4n+2)=\frac{4(4n+1)(8n+7)}{6}\]
First, show that the statement holds for n = 1 (the base case).
After that, assume that it holds for n = k, and under this assumption show that it holds for n = k + 1.

Do you know how to do all that?

Answer 3

I can do n=1 and n=k, just not k+1.. I can't figure that out.

Answer 4

Okay, so having shown the statement holds for n = 1, and assuming the following is true:
\[\small{(4\cdot6)+\cdots+4k(4k+2)+(4k+1)(4(k+1)+2)}\large{=}\small\frac{4(4(k+1)+1)(8(k+1)+7)}{6}\]
Note that the first k terms are the same as the n = k case. In other words,
\[(4\cdot6)+\cdots+4k(4k+2)=\frac{4(4k+1)(8k+7)}{6}\]
Think you can work with that now?

Answer 5

Yes, but n=1 is false.

Answer 6

Oh! Well, in that case, you've shown that the statement is false, and you don't have to do the rest. Unless you're told to prove the statement holds for some other n, like n≥2?

Answer 7

No, just to prove that it is true for all positive integers or to show why it is false. I've got it. Thanks :)

Answer 8

Alright then, you're welcome!