The constant term in the expansion of $$x2+1x2+y+1y8$$ is

Binomial theorem for positive integral Index

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$The constant term in the expansion of {(x^{2} + \frac{1}{x^{2}} + y + \frac{1}{y})}^{8} is$

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$\begin{array}{l} Given expansion {(x^{2} + \frac{1}{x^{2}} + y + \frac{1}{y})}^{8} \\ = \sum \frac{8!}{α! β! γ! δ!} x^{2 α} \frac{1}{x^{2 β}} y^{γ} \frac{1}{y^{δ}} \\ Here α + β + γ + δ = 8 \\ α = β, γ = δ \\ α = 0, β = 0 γ = 4, δ = 4 \\ α = 1, β = 1 γ = 3, δ = 3 \\ α = 2, β = 2 γ = 2, δ = 2 \\ α = 3, β = 3 γ = 1, δ = 1 \\ α = 4, β = 4 γ = 0, δ = 0 \\ Constant term in the expression = 2 (\frac{8!}{0! 0! 4! 4!}) + 2 (\frac{8!}{1! 1! 3! 3!}) + \frac{8!}{2! 2! 2! 2!} = 4200 \end{array}$

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Binomial theorem for positive integral Index

Remember concepts with our Masterclasses.

The constant term in the expansion of x2+1x2+y+1y8 is

Given expansion x2+1x2+y+1y8=∑8!α!β!γ!δ!x2α1x2βyγ1yδ Here α+β+γ+δ=8α=β,γ=δα=0,β=0 γ=4,δ=4α=1,β=1 γ=3,δ=3α=2,β=2 γ=2,δ=2α=3,β=3 γ=1,δ=1α=4,β=4 γ=0,δ=0 Constant term in the expression =28!0!0!4!4!+28!1!1!3!3!+8!2!2!2!2!=4200

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$The constant term in the expansion of {(x^{2} + \frac{1}{x^{2}} + y + \frac{1}{y})}^{8} is$