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 !!
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$\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?
By the binomial expansion, $\displaystyle 4a^{3}=1372$. Solve for a. Understand the pattern. $\displaystyle a^{4}x^{4}+4a^{3}x^{3}+6a^{2}x^{2}+4ax+1$
Thanks, i got it now, i had to equate the 4C3 part of the expansion to 1372x^3 Thanks
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