Q. The coefficients of $x^5$ in the expression $\left(x^2-x-2\right)^5$ is
Binomial Theorem
Solution:
Coefficients of $x^5=\frac{5 !}{a ! b ! c !}\left(x^2\right)^a \cdot(-x)^b(-2)^c$
$=\frac{5 !}{ a ! b ! c !}(-2)^{ c }(-1)^{ b } x ^{2 a + b }$
Now, $a + b + c =5$ and $2 a + b =5$
a
b
c
i
0
5
0
ii
1
3
1
iii
2
1
2
Coefficient $ =\frac{5 !}{0 ! 5 ! 0 !}(-2)^0(-1)^5+\frac{5 !}{1 ! 3 ! 1 !}(-2)^1(-1)^3+\frac{5 !}{2 ! 1 ! 2 !}(-2)^2(-1)^1 $
$ =-1+40-120=-81 $
a | b | c | |
---|---|---|---|
i | 0 | 5 | 0 |
ii | 1 | 3 | 1 |
iii | 2 | 1 | 2 |