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# More Binomial stuff

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In the expantion of (1+3x)^n where n>2 and the coefficient of x^3 in the expansion is ten times the coeffiecnt of x^2, find the value of n.

radio  Oct 5, 2014

#2
+85727
+13

I  just wanted to see if I could get Alan's answer using a different (albeit, more complicated) approach......

So we have

C(n,3)(3x)^3 + C(n,2)(3x)^2

C(n,3)(27x^3) + C(n,2)(9x^2)

27C(n,3)x^3 + 9C(n,2)x^2      and we're told that

27C(n,3) = (10)[(9)C(n,2)]       divide by 9 on both sides

3C(n,3) = 10C(n,2)

3[n!/((n-3)!3!)] = 10[n!/((n-2)!2!)]    divide by n! on both sides

3/[(n-3)!3!)] = 10/[(n-2)!2!]   simplify

(1/2)/(n-3)! = 5/(n-2)!       multiply by 2 on each side

1/(n-3)! = 10/(n-2)!          multiply both sides by (n-2)!

(n-2)!/(n-3)! = 10             the left side simplifies to (n-2)

n-2 = 10

So....n = 12......just as Alan found  !!!

CPhill  Oct 5, 2014
Sort:

#1
+26637
+13

(1 + 3x)n = 1 + n*(3x) + n(n-1)*(3x)2/2! + n(n-1)(n-2)*(3x)3/3! + ...

Ratio of coefficients of x3 term to those of x2 term:  n-2:1  so n-2 = 10.   n = 12.

Check

12(12-1)(12-2)*33/3! = 5940

12(12-1)*32/2! =  594

Alan  Oct 5, 2014
#2
+85727
+13

I  just wanted to see if I could get Alan's answer using a different (albeit, more complicated) approach......

So we have

C(n,3)(3x)^3 + C(n,2)(3x)^2

C(n,3)(27x^3) + C(n,2)(9x^2)

27C(n,3)x^3 + 9C(n,2)x^2      and we're told that

27C(n,3) = (10)[(9)C(n,2)]       divide by 9 on both sides

3C(n,3) = 10C(n,2)

3[n!/((n-3)!3!)] = 10[n!/((n-2)!2!)]    divide by n! on both sides

3/[(n-3)!3!)] = 10/[(n-2)!2!]   simplify

(1/2)/(n-3)! = 5/(n-2)!       multiply by 2 on each side

1/(n-3)! = 10/(n-2)!          multiply both sides by (n-2)!

(n-2)!/(n-3)! = 10             the left side simplifies to (n-2)

n-2 = 10

So....n = 12......just as Alan found  !!!

CPhill  Oct 5, 2014
#3
+965
0

### Yep, I definitely see Grandma’s face :_)

GingerAle  Oct 5, 2014

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