Find coefficient of x^7 doubt ?

Question

Find coefficient of x^7  in expansion of 

(1 + 2x - x^2)^4     

my answer =  -8   by using multinomial  , answer given -1024 

Comments

https://socratic.org/questions/find-the-coefficient-of-x-7-in-the-expansion-of-1-2x-x-2-4

Hope this will help you! Do we have binomial series and all? I don't think so!

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bro  means according to you  answer is -1024  yes this person gave -1024 asnwer =D

Answer 1

Given that  $(1 + 2x - x^{2})^{4}$ 

Let $y=2x-x^{2}$

So we get  $(1+y)^{4}$

Now we can apply Binomial Theorem

$(a+b)^{n} =\sum_{k=0}^{n}\binom{n}{k}(a)^{n-k}\cdot (b)^{k}$

$(1+y)^{4} = \sum_{k=0}^{4}\binom{4}{k}(1)^{4-k} \cdot(y)^{k}$

$(1+y)^{4} = \sum_{k=0}^{4}\binom{4}{k}y^{k}$

we can expend all the term

$(1+y)^{4} = \binom{4}{0}y^{0}+\binom{4}{1}y^{1}+\binom{4}{2}y^{2}+\binom{4}{3}y^{3}+\binom{4}{4}y^{4}$

$(1+y)^{4}=1+4y+6y^{2}+4y^{3}+y^{4}$

Now we can put the valye of $y=2x-x^{2},$and we get

$(1+2x-x^{2})^{4}=1+4(2x-x^{2})+6(2x-x^{2})^{2}+4(2x-x^{2})^{3}+(2x-x^{2})^{4}$

Only one term can be give the $x^{7}$ and term is $(2x-x^{2})^{4}$

So,we can apply again Binomial theorem,

$(2x-x^{2})^{4} =\sum_{r=0}^{4}\binom{4}{r}(2x)^{4-r}\cdot(-x^{2})^{r}$

We can expend all the term

$(2x-x^{2})^{4}=\binom{4}{0}(2x)^{4}\cdot(-x^{2})^{0}+\binom{4}{1}(2x)^{3}\cdot(-x^{2})^{1}+\binom{4}{2}(2x)^{2}\cdot(-x^{2})^{2}+\binom{4}{3}(2x)^{1}\cdot(-x^{2})^{3}+\binom{4}{4}(2x)^{0}\cdot(-x^{2})^{4}$

Only one tern can give $x^{7}$ and the term is  $\binom{4}{3}(2x)^{1}\cdot(-x^{2})^{3}$

$\Rightarrow\binom{4}{3}(2x)^{1}\cdot(-x^{2})^{3}$

$\Rightarrow 4(2x)\cdot(-x^{6})$

$\Rightarrow -8x^{7}$

So,coefficient of $x^{7}$ is$: -8$