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7.Binomial Theorem
normal
The coefficient of $x^8$ in the expansion of $(x-1) (x- 2) (x-3)...............(x-10)$ is :
A
$2640$
B
$1320$
C
$1370$
D
$2740$
Solution
Coeff of $x^{8}$
$=\frac{(1+2+3+\ldots .10)^{2}-\left(1^{2}+2^{2}+\ldots .10^{2}\right)}{2}$
$=1320$
Standard 11
Mathematics