Question

I got most part of the problem..but not yet all of it.. :( it'd be nice if osmeone could help me with this agian..

Answer 1

(2i)^2 = 2n (n+1)(2n+1)/3
a)Show true for n=1
4 = 4
b) Show true for n = k
4 + 16 +36 + 64 + (2k)^2 = 2k (k+1)(2k+1)/3
c) 4 + 16 +36 + 64 + (2k)^2 + (2(k+1))^2 = 2(k+1)((k+1)+1)(2(k+1)+1)/3
2k (k+1)(2k +1)/3 + (2(k+1))^2 = 2(k+1)((k+1)+1)(2(k+1)+1)/3
2k (k+1)(2k+1)/3 + (2(k+1))^2 (3)/3 =
I got this at the end:
4k^3 + 18k^2 +26k + 12/3 (which is correct.. but I dont know where to go from here..)

Answer 2

2i^2 ?

Answer 3

(2i)^2

Answer 4

Is always -4

Answer 5

i stands for n. not imaginary

Answer 6

OK, (2n)^2 and you want an induction proof, right?

Answer 7

um yeah..like the way i showed it..kidna..but i got stuck at the end.. :(

Answer 8

\[\begin{align}
\dfrac{2k (k+1)(2k +1)}{3} + (2(k+1))^2 &= \dfrac{2(k+1)((k+1)+1)(2(k+1)+1)}{3}\\
\dfrac{(2k^2+2k)(2k +1)}{3} + (2k+2)^2 &= \dfrac{2(k+1)(k+2)(2k+3)}{3}\\
\dfrac{4k^3+6k^2+2k}{3} + \dfrac{3(2k+2)^2}{3} &= \dfrac{(2k^2+6k+4)(2k+3)}{3}\\
\dfrac{4k^3+6k^2+2k}{3} + \dfrac{3(4k^2+8k+4)}{3} &= \dfrac{4k^3+18k^2+26k+12}{3}\\
\dfrac{4k^3+6k^2+2k}{3} + \dfrac{12k^2+24k+12}{3} &= \dfrac{4k^3+18k^2+26k+12}{3}\\
\dfrac{4k^3+18k^2+26k+12}{3} &= \dfrac{4k^3+18k^2+26k+12}{3}
\end{align}\]

Answer 9

whoaa.

Answer 10

Assume true for k, does that imply true for k+1

Answer 11

Oh..nbouscal..we'arent allowd to touch the right side..

Answer 12

Yes it does..

Answer 13

Then you are done.

Answer 14

um...u have to prove it alegbraically..

Answer 15

Bah. You can always touch the right side. Anyway, just do the inverse of anything I did to the right side to the left side instead and then you're good.

Answer 16

okay lemme try.. (lol its my teachers pet peeve on test to only mess with the left side and not the rite.)

Answer 17

Your teacher's an idiot, then. (in my humble opinion)

Answer 18

*Best Answer*^

Answer 19

Im sorry if i m bothering you..but i tried it backwards, and tried doing synthetic division, firstdividng it by k+1 to get 4k^2 +14k + 12 and divind that by k+2, to get4k +6..

which:
(k+1)(k+2)(4k+6)
(k+1) ((k+1)+1)(2(2k+3)/3..... and not what i needed to get.. :( sorry!1 :(

Answer 20

\[
\begin{align}
\dfrac{2k (k+1)(2k +1)}{3} + (2(k+1))^2 &= \dfrac{2(k+1)((k+1)+1)(2(k+1)+1)}{3}\\
\dfrac{(2k^2+2k)(2k +1)}{3} + (2k+2)^2 &= \\
\dfrac{4k^3+6k^2+2k}{3} + \dfrac{3(2k+2)^2}{3} &=\\
\dfrac{4k^3+6k^2+2k}{3} + \dfrac{3(4k^2+8k+4)}{3} &= \\
\dfrac{4k^3+6k^2+2k}{3} + \dfrac{12k^2+24k+12}{3} &= \\
\dfrac{4k^3+18k^2+26k+12}{3} &=\\
\dfrac{(2k^2+6k+4)(2k+3)}{3}&=\\
\dfrac{2(k+1)(k+2)(2k+3)}{3}&=\\
\dfrac{2(k+1)((k+1)+1)(2(k+1)+1)}{3}&=\dfrac{2(k+1)((k+1)+1)(2(k+1)+1)}{3}\\
\end{align}
\]

Answer 21

Um...one last question.. : How'd you get 2k^2 + 6k + 4 and 2k+3?