# In the expansion of (3+x/2)^n where n is a positive integer, find the value of n if the ratio of coefficient of x^2 to the coefficient of x^4 is 6

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### 1 Answer

Find n if the ratio of the coefficient on x^2 to the coefficient on x^4 is 6 for the expansion of `(3+x/2)^n` :

The expansion of `(3+x/2)^n` is:

`3^n+([n],[1])3^(n-1)(x/2)+...+([n],[n-4])1/2^4x^4 3^(n-4)`

`+([n],[n-3])1/2^3 x^3 3^(n-3)+([n],[n-2])1/2^2 x^2 3^(n-2)+n1/2 x 3^(n-1)+3^n`

The coefficient of the `x^2` term is `([n],[n-2])1/4 3^(n-2)` and the coefficient of the `x^4` term is `([n],[n-4])1/16 3^(n-4)`

The required ratio is `(([n],[n-2])1/4 3^(n-2))/(([n],[n-4]) 1/16 3^(n-4))`

`(1/4)/(1/16)=4` and `(3^(n-2))/(3^(n-4))=3^(n-2-(n-4))=3^2=9` so rewriting and setting equal to 6 we get:

`36(([n],[n-2]))/(([n],[n-4]))=6==>(([n],[n-2]))/(([n],[n-4]))=1/6`

Use the definition for a combination:

`((n(n-1)(n-2)!)/((n-2)!2!))/((n(n-1)(n-2)(n-3)(n-4)!)/((n-4)!4!))=1/6` a lot of stuff cancels

`==>12/((n-2)(n-3))=1/6`

`==>(n-2)(n-3)=72`

`==>n^2-5n-66=0`

`==>(n-11)(n+6)=0` and since n>0

**n=11 is the solution.**

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