7.Binomial Theorem
hard

$\left(1+x+x^{2}\right)^{10}$ के प्रसार में $x^{4}$ का गुणांक है

A

$615$

B

$625$

C

$595$

D

$575$

(JEE MAIN-2020)

Solution

$\left(1+x+x^{2}\right)^{10}$

$=^{10} \mathrm{C}_{0}+^{10} \mathrm{C}_{1} \mathrm{x}(1+\mathrm{x})+^{10} \mathrm{C}_{2} \mathrm{x}^{2}(1+\mathrm{x})^{2}$

$+^{10} \mathrm{C}_{3} \mathrm{x}^{3}(1+\mathrm{x})^{3}+^{10} \mathrm{C}_{4} \mathrm{x}^{4}(1+\mathrm{x})^{4}+\ldots \ldots$

Coeff. of $\mathrm{x}^{4}=^{10} \mathrm{C}_{2}+^{10} \mathrm{C}_{3} \times^{3} \mathrm{C}_{1}+^{10} \mathrm{C}_{4}=615$

Standard 11
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.