In the binomial expansion of #(1+ax)^n#, where #a# and #n# are constants, the coefficient of #x# is 15. The coefficient of #x^2# and of #x^3# are equal. What is the value of #a# and of #n#?

1 Answer
Jacob
Nov 26, 2017

#a=6, n=2.5#

Explanation:

The formula for the binomial expansion of #(1+ax)^n# is:
#1+n(ax)+(n*(n-1))/(2!) (ax)^2...(n(n-1)...(n-r+1))/(r!) (ax)^r#

Therefore the #x^1# coefficient is #an=15#
If the #x^2# and #x^3# coefficients are equal, this must mean that:
#(n(n-1))/(2!)(a)^2=(n(n-1)(n-2))/(3!)(a)^3#

Taking out factors of #(n(n-1))/2 a^2# gives: #1=(n-2)/3a#

which expands to: #3=an-2a#

We know that #an=15#, so #3=15-2a# which rearranges to #2a=12#, so #a=6#.

#an=15# so #n=15/a=15/6=2.5#

You can check the answer by putting it back into the original formula.

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