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#