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

[DVDVDVDV]

Title

 

[Patato]

set up two equations from part ii)

one let x = 1, the other let x = -1
add, they cancel
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]

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]

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

 

[brachester]

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]

^ hahah I tried to do induction didn't work

 

[D94]

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

 

[someth1ng]

Yeah, fuck that. LOL.

 

[cutemouse]

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