+354 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.
+129899 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 !!!
+33661 (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
+129899 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 !!!
