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  • The absolute value of coefficient of x3 in the expansion of <img alt="\left(1-x+x^2\right)^5" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cleft%281-x&plus;x%5E2%5Cright%29

The absolute value of coefficient of x3 in the expansion of \left(1-x+x^2\right)^5 is

Option: 1

-25


Option: 2

28


Option: 3

-30


Option: 4

30


Answers (1)

best_answer

\text{The general term in the expansion of}\ \left(1-x+x^2\right)^5\ is\\ \begin{aligned} & \frac{5 !}{r ! s ! t !}(1)^r(-x)^s\left(x^2\right)^t, \text { where } r+s+t=5 \\ & =\frac{5 !}{r ! s ! t !}(-1)^s x^{s+2 t}, \text { where } r+s+t=5 \end{aligned}\\ \text{For this term to contain}\ x^3,\ \text{we must have}\ s+2 t=3\\ \begin{gathered} \text { But, } r+s+t=5 \\ \therefore \quad s=3-2 t, r=2+t \\ \text { Now, } t=0 \Rightarrow r=2, s=3 \\ t=1 \Rightarrow r=3, s=1 \end{gathered}\\ \text{Thus, there are two terms containing}\ x^3.\\ \text{Putting $r=2, s=3, t=0$ and $r=3, s=1, t=1$ respectively in (i), we obtain that the terms are}\\ \frac{5 !}{2 ! 3 ! 0 !}(-1)^3 x^3 \text { and } \frac{5 !}{3 ! 1 ! 1 !}(-1) x^3

\text { Hence, coefficient of } x^3=-\frac{5 !}{2 ! 3 !}-\frac{5 !}{3 !}=-10-20=-30

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