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Finding value independent of x in a binomial expansion?



A = 420
k = 2


These were worked out from the earlier part (a)

What is the method of calculating the independent term ? Someone plz help :smile:

Thanks
Reply 1
Avatar for nota bene
nota bene
15
Find for r=5 (this I did by recognition and some thought...dont really think there is a 'method')

edit: So

Spoiler

Reply 2
Avatar for coffeym
coffeym
10
15C5(x10)(2x2)5=96096^{15}C_5(x^{10})(\frac{2}{x^2})^5 = 96096

I agree with the above post too, just find an r value so that the xx terms cancel.
Reply 3
Avatar for Mos Def
Mos Def
OP
17
what is the r? :s-smilie:
(x+kx2)15=∑r=015(15r)x15−r(kx2)r(x+\frac{k}{x^2})^{15} = \sum_{r=0}^{15} {15 \choose r}x^{15-r}(\frac{k}{x^2})^r
that is, the fomrula for a binomial expansion, putting n=15

So the general term is (15r)x15−r(kx2)r=(15r)x15−r(kx−2)r{15\choose r}x^{15-r}(\frac{k}{x^2})^r = {15\choose r}x^{15-r}(kx^{-2})^r

Can you find a value of r such that the power of x is zero? (Get the x indices together and solve the equation, or just do it by inspection). Evaluate the general term for this value of r.
nCr. Number of ways of choosing r objects from n items.

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