Binomial Question (1 Viewer)

[Michaelmoo]

In the expancsion of (3+4x)^n, the coefficients of x^2 and x^3 are in the ratio of 3:4. Find the value of n.

I cant seem to get to the write answer. Can anyone help?

Thanks.

 

[cookie.banana]

i'd say

7:6 5'4 185654>e=mc2

 

[Michaelmoo]

coefficient of x^2
= nC2 . 3^(n-2) . (4)^2
= [n! . 3^(n-2) . (4)^2] / [2! . (n-2)!]

coefficient of x^3
= nC3 . 3^(n-3) . (4)^3
= [n! . 3^(n-3) . (4)^3] / [3! . (n-3)!]

{[n! . 3^(n-2) . (4)^2] / [2! . (n-2)!]} / {[n! . 3^(n-3) . (4)^3] / [3! . (n-3)!]} = 3/4

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

18 / [8 . (n-2)] = 3/4

n = 5
quote]

Yep. Thanks. Thought the answers might be wrong.

 

[renny 123]

I posted this same question up yesterday
I got 5 aswell and couldn't figure out what i did wrong.
The answer in the book is incorrect!