Cengage algebra jee advanced.

Question

Coefficient of x^8 in ( (1-x^6)/(1-x) )^3.

Comments

$[x^8] (1- x^6)^3 (1 – x)^{-3}$

$[x^8] \sum_{r=0}^{3} \binom{3}{r}(-1)^r x^{6r} \sum_{k=0}^{\infty} \binom{3+k-1}{k} x^{k}$

So,  $r=0,k=8$ and $r=1,k=2$

Hence, Answer is $\binom{3}{0}\binom{3+8-1}{8} - \binom{3}{1}\binom{3+2-1}{2} = 27$

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Thankyou.

Answer 1

Given ,

$\large (\frac{1-x^{6}}{1-x})^{3}$

=$\large (1-x^{6})^{3}(1-x)^{-3}$

=$\large (1-3x^{6}+3x^{12}-x^{18})(1-x)^{-3}$

Now we need to expand $\large (1-x)^{-3}$.

$\large (1-x)^{-3}$                [By extended binomial theorem]

=$\large (1+\binom{-3}{1}(-x)+\binom{-3}{2}x^{2}+\binom{-3}{3}(-x)^{3}+.......)$

Now, $\large \binom{-n}{r}=(-1)^{r}\binom{n+r-1}{r}$  [only if n is integer ]

=$\large (1+3x+6x^{2}+10x^{3}+15x^{4}+21x^{5}+28x^{6}+36x^{7}+45x^{8}+........)$

 

So putting this value in the expression ,

=$\large (1-3x^{6}+3x^{12}-x^{18})$ . $\large (1+3x+6x^{2}+10x^{3}+15x^{4}+21x^{5}+28x^{6}+36x^{7}+45x^{8}+........)$

So Coefficient of $\large x^{8}$ is $(45+(-3)*6)=27$ .

Comments

you might missed $(-1)^r$ for formula of $\binom{-n}{r}$

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@ankitgupta.1729 yes sir missed that by mistake edited 😬

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Thankyou