How to solve this binomial expansion? (1 Viewer)

[rhatc04]

Thanks

 

[Hamburgler]

Consider double sigma notation, collating like x terms into one single power

 

[rhatc04]

Consider double sigma notation, collating like x terms into one single power

I've collected all the x terms into single power to get x^10-4k and the coefficient is 7C2 x 3C2 however it is incorrect

 

[Deem_Skills]

i think u have to do it manually

 

[Deem_Skills]

if u use sigma i got this



which isnt right

 

[rhatc04]

i think u have to do it manually

View attachment 36451

Yes that's correct, thanks so much

 

[rhatc04]

if u use sigma i got this

View attachment 36452

which isnt right

Yeah that's what I did and I got the same thing

 

[Hamburgler]

if u use sigma i got this

View attachment 36452

which isnt right

Yeah that's what I did and I got the same thing

if u use sigma i got this

View attachment 36452

which isnt right

You did it wrong, you can’t use same k values u have to use different ones. Consider what I have done.

 

[Drongoski]

This is how I would do it.



Not very different from earlier efforts above.

 

[rhatc04]

I'm not too sure how to approach this question, any tips?

 

[Hamburgler]

Apologies for lots of scribbles.

 

[rhatc04]

In this step, why is it n-i and j-n ?

Shouldn't it be (-x)^i and x^-j

 

[Hamburgler]

View attachment 36458

In this step, why is it n-i and j-n ?

Shouldn't it be (-x)^i and x^-j

Because both are symmetrical. (1+x)^n = nCr * 1^r * x^n-r = nCr * 1^n-r * x^r
If you stick with one, u have to use the same one i.e. can't use one this way and the other that way

 

[rhatc04]

Because both are symmetrical. (1+x)^n = nCr * 1^r * x^n-r = nCr * 1^n-r * x^r
If you stick with one, u have to use the same one i.e. can't use one this way and the other that way

I think I get it now, thanks! Your method of using i and j is really efficient