- Home
- Standard 11
- Mathematics
7.Binomial Theorem
easy
${\left( {{x^2} - 2x} \right)^{10}}$ ના વિસ્તરણમાં ${x^{16}}$ નો સહગુણક મેળવો.
A
$-1680$
B
$1680$
C
$3360$
D
$6720$
Solution
(c) The coefficient of ${x^{16}}$ in the expansion of ${({x^2} – 2x)^{10}}$
= The coefficient of ${x^{16}}$ in ${x^{10}}{(x – 2)^{10}}$
= The coefficient of ${x^6}$ in ${(x – 2)^{10}}$
= $^{10}{C_4}{.2^4},\,\,\,\,\left( {\,\,{T_{r + 1}} = {\,^n}{C_r}{x^{n – r}}{a^r}} \right)$
$ = 210 \times 16 $
$= 3360$.
Standard 11
Mathematics
