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in Binomial Theorem by (25.0k points)
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If (1 – x + x2)n = a0 + a1x + a2x2 + … + a2nx2n, find the value of a0 + a2 + a4 + … + a2n.

1 Answer

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Best answer

(1 – x + x2)n = a0 + a1x + a2x2 + … + a2nx2n

At x = 1,

(1 – 1 + 12)n = a0 + a1(1) + a2(1)2 + … + a2n(1)2n 

a0 + a1 + a2 + … + a2n = 1 …(1)

At x = -1, 

(1 – (-1) + (-1)2)n = a0 + a1(-1) + a2(-1)2 + … + a2n(-1)2n 

a0 - a1 + a2 - … + a2n = 3n …(2) 

On adding eq.1 and eq.2 

(a0 + a1 + a2 + … + a2n) + (a0 - a1 + a2 - … + a2n) = 1 + 3n 

2(a0 + a2 + a4 + … + a2n) = 1 + 3n 

a0 + a2 + a4 + … + a2n\(\frac{1+3^n}{2}\)

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