Question #00978

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Answer 1 · 2017-01-10T00:50:01

See below.

$(1+ax)^n=1+nax+(n(n-1))/(2!)(ax)^2+(n(n-1)(n-2))/(3!)(ax)^3+cdots$

The coefficient of $x$ is $na$ and also

$a^3(n(n-1)(n-2))/(3!)=6a^2(n(n-1))/(2!)$

so the conditions are

${(n a=12),(a(n-2)/(3!)=6/(2!)):}$

solving for $a,n$ we have

$n = -4$ and $a = -3$ or

$1/(1-3x)^4$

and the expansion is valid for $x ne 1/3$