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  4. The co-efficient of x3 in the expansion of (1-x+x2)5 is

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Q. The co-efficient of $x^3$ in the expansion of $(1-x+x^2)^5$ is

Binomial Theorem

Solution:

No. of terms in
$\left(1 - x+ x^{2}\right)^{5} = 1 + x\left(x- 1\right)^{5}$
$= \,{}^{5}C_{0} + \,{}^{5}C_{1}x\left(x-1\right)+ \,{}^{5}C_{2}x^{2}\left(x-1\right)^{2}$
$+ \,{}^{5}C_{3}x^{3} \left(x-1\right)^{3}+ \,{}^{5}C_{4}x^{4}\left(x-1\right)^{4}+ \,{}^{5}C_{5}x^{5}\left(x-1\right)^{5}$
$\therefore $ co - eff. $x^{3} = - 2\cdot^{5}C_{2}- \,{}^{5}C_{3} = -30$