I hope this is the right part of the forum for this question.

In the expansion of (1+ax)^4, the coefficient of x^3 is 1372. Find the constant a.

Please Help !! (Crying)

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- Oct 2nd 2009, 12:44 PMcreatively12Binomial Theorem or Binomial Coefficient
I hope this is the right part of the forum for this question.

In the expansion of (1+ax)^4, the coefficient of x^3 is 1372. Find the constant a.

Please Help !! (Crying) - Oct 2nd 2009, 12:54 PMDefunkt
$\displaystyle (1+ax)^4 = \sum_{n=1}^{4} \binom{4}{n}a^nx^n $

So the coefficient of x^3 is $\displaystyle \binom{4}{3}a^3$

Can you solve this now? - Oct 2nd 2009, 12:55 PMgalactus
By the binomial expansion, $\displaystyle 4a^{3}=1372$.

Solve for a.

Understand the pattern. $\displaystyle a^{4}x^{4}+4a^{3}x^{3}+6a^{2}x^{2}+4ax+1$ - Oct 2nd 2009, 01:06 PMcreatively12
Thanks, i got it now, i had to equate the 4C3 part of the expansion to 1372x^3

Thanks