miguel_fugdey
Member
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?
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indipendant variables in binomials (1 Viewer)
- Thread starter miguel_fugdey
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SoFTuaRiaL
Active Member
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 ....
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
Member
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?)
(x2 + 3/x)6
(that is, the 2 and the 6 are powers) (how do you do that by the way?)
SoFTuaRiaL
Active Member
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
Member
sweeeet thanks da mon...it really had me stumped...damned binomial theorum is so the hardest 3u topic