2
$\begingroup$

$$(1-x^3)^{n} = \sum_{r=0}^n a_{r}x^{r}(1-x)^{3n-2r}$$ where n belongs to the set of natural numbers.

Clearly, it should be solved using Binomial theorem. But i don't seem to find out how. Please help me by providing a solution and do mention the steps. Thanks.

$\endgroup$
1
$\begingroup$

HINT:

For $x\ne1,$ $$\dfrac{1-x^3}{(1-x)^3}=\dfrac{1+x+x^2}{(1-x)^2}=\dfrac{(1-x)^2+3x}{(1-x)^2}=1+3\cdot\dfrac x{(1-x)^2}$$

$$\implies\left(\dfrac{1-x^3}{(1-x)^3}\right)^n=\left(1+3\cdot\dfrac x{(1-x)^2}\right)^n$$

Now look at the Binomial series

$\endgroup$
  • $\begingroup$ I can see it now. So the answer comes out as C(n, r). 3^r? $\endgroup$ – Phill2 Feb 23 '16 at 16:07
  • $\begingroup$ @Phill2, Rightly so $\endgroup$ – lab bhattacharjee Feb 23 '16 at 16:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.