Binomial theorem for positive integral Index

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Question

The coefficient of $x^{4}$ in the expansion of ${(1 + x + x^{2})}^{10}$ is__________

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Solution

$General term = \frac{10!}{α! β! γ!} \times x^{β + 2 γ}, where α + β + γ = 10$
$For coefficient of x^{4}, β + 2 γ = 4$
So, the possible sets will be
$\begin{array}{l} γ = 0, β = 4, α = 6 \Rightarrow \frac{10!}{6! 4! 0!} = 210 \\ \begin{array}{l} γ = 1, β = 2, α = 7 & \Rightarrow \frac{10!}{7! 2! 1!} = 360 \\ γ = 2, β = 0, α = 8 \Rightarrow & \frac{10!}{8! 0! 2!} = 45 \end{array} \end{array}$
Total = 615

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Binomial theorem for positive integral Index

Remember concepts with our Masterclasses.

The coefficient of x4in the expansion of (1+x+x2)10 is__________

General term =10!α!β!γ!×xβ+2γ, where α+β+γ=10 For coefficient of x4,β+2γ=4So, the possible sets will beγ=0,β=4,α=6 ⇒ 10!6!4!0!=210γ=1,β=2,α=7⇒10!7!2!1!=360 γ=2,β=0,α=8⇒10!8!0!2!=45 Total = 615

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The coefficient of $x^{4}$ in the expansion of ${(1 + x + x^{2})}^{10}$ is__________