Question

Please Help!

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

1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = n(6n^2 -3n -1) /2

I got as far as proving that the formula holds true for n=1. I need help with the rest please!

Answer 1

@sami-21

Answer 2

replace \(n\) by \(k\) and assert that
\[1^2 + 4^2 + 7^2 + ... + (3k - 2)^2 = \frac{k(6k^2 -3k -1)}{2})\] is true, then prove the formula for \(k+1\)

Answer 3

Right I'm sorry I forgot to say I did that too! I am having trouble proving formula for k+1

Answer 4

the \(k+1\) statement is
\[1^2 + 4^2 + 7^2 + ... + (3k - 2)^2 +(3(k+1)-2)^2\]
\[= \frac{(k+1)(6(k+1)^2 -3(k+1) -1)}{2}\]

Answer 5

and you can claim "by induction" that the left hand side is equal to
\[\frac{k(6k^2 -3k -1)}{2}+(3(k+1)-2)^2\] and it is really ugly algebra from here on in

Answer 6

so from that last do I try to FOIL?

Answer 7

multiply out, combine like terms etc etc it is a pain
i can't do it here, but according to wolfram it is
\[1/2 (6 k^3+15 k^2+11 k+2)\]

Answer 8

here is the left hand side of the equal sign
http://www.wolframalpha.com/input/?i= \frac{k%286k^2+-3k+-1%29}{2}%2B%283%28k%2B1%29-2%29^2

Answer 9

here is the right hand side
http://www.wolframalpha.com/input/?i=%3D \frac{%28k%2B1%29%286%28k%2B1%29^2+-3%28k%2B1%29+-1%29}{2}

Answer 10

That made my head hurt... so what do I do after? because it looks nothing like (k+1)

Answer 11

you can see that they are the same

Answer 12

give me one sec

Answer 13

copy and paste, the link will not work as it is

Answer 14

at least you will know what you are working towards

Answer 15

they are the same just simplified right? but I thought they both had to equal (k+1)

Answer 16

oh hell no

Answer 17

what you get on the right is the formula replacing \(k\) by \(k+1\)

Answer 18

i just copied and pasted, then replaced all the \(k\)'s i saw by \(k+1\)

Answer 19

the left hand side is the sum from 1 to \(k+1\) instead of the sum from one two k

Answer 20

good luck with the algebra, it looks punishing, but it is just algebra, and now you know what you should get when you do it

Answer 21

Ok thanks!

Answer 22

Could you help me with another? Or should I post it separate?