indipendant variables in binomials (1 Viewer)

[miguel_fugdey]

i just had an exam, and it had a question about indipendant variables in a binomial expansion...i figured this would occur when , in nCra~(n-r)b~r (general binomial expansion), n=r, but i got the answer wrong...any clues?

 

[kini mini]

Sorry, I don't understand the question. Can you type the whole thing up?

 

[SoFTuaRiaL]

say u got (ax + b)^n ..........
expansion is E from r=0->n nCr (ax)^n-r b^r ..... i think

independent of variable is when power of x is 0, in this case, n-r=0, ie, n=r. so, just substitute n=r
in other cases, where say hte expression is (ax^2 + bx^4), the n-r wud be something else ....

 

[miguel_fugdey]

Find the indipendant variable int he binomial expansion:

(x2 + 3/x)6

(that is, the 2 and the 6 are powers) (how do you do that by the way?)

 

[SoFTuaRiaL]

Originally posted by miguel_fugdey
Find the indipendant variable int he binomial expansion:

(x2 + 3/x)6

(that is, the 2 and the 6 are powers) (how do you do that by the way?)

exp : (x^2 + 3/x)^6
therefore, expansion is E from r->6 of 6Cr * (x^2)^6-r * (3/x)^r
which is E r->6 6Cr * x^(12-2r) * (3/x)^r
bringing the (1/x)^r together with the other one, we get x^(12-3r).
now, we need this to be 0. 12-3r=0, so r=4.
sub r=4 in the rest of them, answer is 6C4 * 3^4 (whatever that turns out to be). i think its 1215


btw, E stands for sigma

 

[miguel_fugdey]

sweeeet thanks da mon...it really had me stumped...damned binomial theorum is so the hardest 3u topic