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JEE Main20253 Apr 2025Evening ShiftMathematicsBinomial TheoremActual

Let (1+x+x^2 )¹⁰=a₀+a₁ x+a₂ x^2+ .+a₂₀ x²⁰ . If (a₁+a₃+a₅+ .+a₁₉ )-11 a ₂=121 k , then k is equal to .

Correct answer

0

Step-by-step solution

(1+x+x^2 )¹⁰=a₀+a₁ x+a₂ x^2+ .+a₂₀ x²⁰ 3¹⁰= a ₀+ a ₁+ a ₂+ .+ a ₂₀ ...(i) 1=a₀-a₁+a₂ . .+a₂₀ ...(ii) (i) - (ii) a₁+a₃+ .+a₁₉= 3¹⁰-1 2 =29524 aligned Also & 1+ x (1+ x ) ¹⁰=1 & + ¹⁰ C ₁ x (1+ x )+ ¹⁰ C ₂ x ^2(1+ x )^2+ . aligned a ₂= ¹⁰ C ₁+ ¹⁰ C ₂=55 ( a ₁+ a ₃+ + a ₁₉ )-11 a ₂ 121 =239

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