I've spent a long time trying to solve these few questions, but I still couldn't get them, just 14c and 15. Pls help thanks
View attachment 36729
for (b),
expand (1+x)^2n:
2nC0 + 2nC1x + 2nC2x^2 + ... + 2nC(n-r)x^n-r + .. + 2nC2(n)x^n
so coeff is 2nC(n-r)
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Construct the expansion of both (1+x)^n in a form where you can take out the coeff of x^n-r directly
(1+x)^n = nC0(x^0) + nC1(x)^1 + ... + nC(n-r)x^n-r + nCnx^n
(1+x)^n = nCr(x^n-r) + nC(r+1)x^n-r-1 + .. + nC1(x^0)+ nC0(x^-r)
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= nC0*nCr(x^n-r) + nC1nC(r+1)(x^n-r) + ... + nC0*nCn)(x^n-r)
= f(r)
(1+x)^2n = (1+x)^n * (1+x)^n
so f(r) = 2nCn-r when considering coeffs of x^n-r
(This is for anyone else seeing who doesnt know b, not u)
(c)
Consider first n terms for all expansions only
(1+x)^3n = 3nC0 + ... + 3nCn(x^n)
Looking for coefficient of 3nCn only
(1+x)^2n = 2nCn(x^n) +2nCn-1(x^n-1) + ... + 2nC0x^0
(1+x)^n = nC0(x^0) + nC1(x^1) + ... + nCn(x^n)
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= 2nCn*nC0 + ... + nCn *2nC0(x^n)
= nC0*f(0)+ ... + nCnf(n)
as (1+x)^3n = (1+x)^2n * (1+x)^n
nC0*f(0) + ... + nCnf(n) = 3nCn
15
Sorry if I'm wrong but can't you just make x = 1? so the whole sum = 1 or am i missing something
