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7.Binomial Theorem
hard
$(1- x )^{101}\left( x ^{2}+ x +1\right)^{100}$ के प्रसार में $x ^{256}$ का गुणांक है
A
${-}^{100} \mathrm{C}_{16}$
B
$^{100} \mathrm{C}_{16}$
C
$^{100} \mathrm{C}_{15}$
D
$-{ }^{100} \mathrm{C}_{15}$
(JEE MAIN-2021)
Solution
$y=(1-x)(1-x)^{100}\left(x^{2}+x+1\right)^{100}$
$y=(1-x)\left(x^{3}-1\right)^{100}$
$y=\left(x^{3}-1\right)^{100}-x\left(x^{3}-1\right)^{100}$
Coff. Of $x^{256}$ in $y=-$ coff of $x^{255}$ in $\left(x^{3}-1\right)^{100}$
$={ }^{-100} \mathrm{C}_{85}(-1)^{15}={ }^{100} \mathrm{C}_{15}$
Standard 11
Mathematics