Question #00978

Question

Question #00978

1 Answer

Impact of this question

1338 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-10T00:50:01

See below.

Explanation:

${(1 + a x)}^{n} = 1 + n a x + \frac{n (n - 1)}{2!} {(a x)}^{2} + \frac{n (n - 1) (n - 2)}{3!} {(a x)}^{3} + \dots$ $(1+ax)^n=1+nax+(n(n-1))/(2!)(ax)^2+(n(n-1)(n-2))/(3!)(ax)^3+cdots$

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

$a^{3} \frac{n (n - 1) (n - 2)}{3!} = 6 a^{2} \frac{n (n - 1)}{2!}$ $a^3(n(n-1)(n-2))/(3!)=6a^2(n(n-1))/(2!)$

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 $x ne 1/3$

Science

Math

Humanities

... and beyond

Question #00978

1 Answer

Explanation:

Related questions

Impact of this question