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**+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

#1**+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 x^{3} term to those of x^{2} term: n-2:1 so n-2 = 10. n = 12.

Check

12(12-1)(12-2)*3^{3}/3! = 5940

12(12-1)*3^{2}/2! = 594

Alan
Oct 5, 2014

#2**+13 **

Best Answer

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