Question

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

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

Answer 1

to get the summation u do :
sigma(r=1 to n)[4r( 4r + 2)] , ans would be a cubic polynomial but you have given ans as square polynomial so its false.
OR u can check using values as n=1,2,3... and check its wrong.

2)sigma(1 to n)[(3r-2)^2] ans would be a cubic polynomial and the given ans is also so we need to check.
=sigma(1 to n)(9r^2-12r+4)
=[9*n(n+1)(2n+1)/6-12*n(n+1)/2+4n] {using sigma(r^2)=n(n+1)(2n+1)/6 ; sigma(r)=n(n+1)/2 sigma(1)=n}
=n(6n^2-3n-1)/2

Answer 2

I'm not really sure what you did, could you explain?

Answer 3

\[(4\cdot6)+(5\cdot7)+(6\cdot8)+\cdots+4n(4n+2)=\frac{4(4n+1)(8n+7)}{6}\]
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 4

Not really, I somewhat understand n =1 but when I tried working it out I wasn't getting the correct answer.

Answer 5

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 6

So for n = k you just swap the letter n out for k?

Answer 7

Yes

Answer 8

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 9

Do you get it now?

Answer 10

Yes, that makes sense. Thank you so much!

Answer 11

You're welcome! :)