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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