Binomial theorem (1 Viewer)

[bboyelement]

cant seem to get this question

in the expansion of (2+3x)^n the coefficients of the x^3 and x^4 are in the ratio 8:15. Find n

 

[word.]

(2+3x)^n = 2^n + nC1*2^(n-1)*(3x) + nC2*2^(n-2)*(3x)^2 + nC3*2^(n-3)*(3x)^3 + nC4*2^(n-4)*(3x)^4 + ...

nC3*2^(n-3)*27
--------------------- = 8/15
nC4*2^(n-4)*81

nC3/nC4 * 2^(n-3)/2^(n-4) * 27/81 = 8/15

nC3 = n!/[3!(n-3)!]
nC4 = n!/[4!(n-4)!]
nC3/nC4 = 4/(n-3)

2^(n-3)/2^(n-4) = 2^[(n-3)-(n-4)] = 2

27/81 = 1/3

4/(n-3) * 2 * 1/3 = 8/15

8/3 * 1/(n-3) = 8/15

n = 8

 

[bboyelement]

thanks a lot