+0  
 
0
427
3
avatar+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.

radio  Oct 5, 2014

Best Answer 

 #2
avatar+90023 
+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
avatar+27052 
+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
avatar+90023 
+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
 #3
avatar+1184 
0

This looks like you just tested your newly sharpened machete.

Charlotte is right, CPhill, you are $&#**^% amazing!

 

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

GingerAle  Oct 5, 2014

13 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.