# How do you do the last question? (1 Viewer)

Title

#### Patato

##### New Member
set up two equations from part ii)

one let x = 1, the other let x = -1
you get 2 ( the middle part) = some shit
divide both sides by 2.

if i recall correctly

sorry about the brievity lol i dont understand latex and i dont have the paper with me

#### DVDVDVDV

Here's the question

Assume that n is even. Shaw that, for n>=4,

(nC2)*2^2 + (nC4)4^2 + (nC6)6^2.... + (nCn)n^2 = n(n + 1)2^(n-3)

#### brettymaccc

##### Member
Here's the question

Assume that n is even. Shaw that, for n>=4,

(nC2)*2^2 + (nC4)4^2 + (nC6)6^2.... + (nCn)n^2 = n(n + 1)2^(n-3)
Induction?

Ie. Prove for n = 4, assume for n = k, where k is an even integer >= 4, then prove for n = k + 2?

#### Patato

##### New Member
fingers crossed i didnt stuff up latex...

this is how i did it, hopign its all right

you basically just set up 2 equations, one hwere x = 1, the other x = -1, add the two together and the nc1, nc3,...nCn-1's all cancel

$\bg_white \textup{We know that}\\ nC1 + nC2.2^{2} + ... + nCn^{2} = n(n+1).2^{n-2} .... (1) \textup{ (from ii)}\\\\ \textup{Now using (i)}\\ \sum nCr.r.x^{r} = nx(1+x)^{n-1}\\\\ \textup{So let x = -1 in that equation}\\\\ \textup{We get...}\\\\ -nC1 + nC2.2^{2} - ... + nCn^{2} = 0 .... (2)\\\\ \textup{Adding (1) and (2)}\\\\ 2(nC2.2^{2} + nC4.4^{2} + ... + nCn.n^{2})) = n(n+1).2^{n-2}\\\\ \textup{Divide both sides by 2}\\\\ nC2.2^{2} + nC4.4^{2} + ... + nCn.n^{2})) = n(n+1).2^{n-3}$

#### brachester

##### Member
Induction?

Ie. Prove for n = 4, assume for n = k, where k is an even integer >= 4, then prove for n = k + 2?
lolololol, that's exactly what i was trying to do even though i'm not supposed to when they didn't ask for induction. It didn't work (as expected)

#### eyeswideshut

##### Member
^ hahah I tried to do induction didn't work

#### D94

##### Well-Known Member
Yeah, the above method works well, probably quicker than the way I did it - setting up another expansion with (1-x)n. Sort of coincidental and luck, I did this type of question with my tutor the week before

Last edited:

#### someth1ng

##### Retired Nov '14
Yeah, fuck that. LOL.

#### cutemouse

##### Account Closed
n=2k where k is an integer should be a starting point...