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in Binomial Theorem by (23.7k points)
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If (1 – x + x2)n = a0 + a1x + a2x2 + ...... + a2nx2n, then a0 + a2 + a4 + ..... + a2n equals

(a) \(\frac{3^n -1}{2}\)

(b) \(\frac{1-3^n}{2}\)

(c) 3n + \(\frac{1}{2}\) 

(d) \(\frac{3^n +1}{2}\)

1 Answer

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Answer : (d) \(\frac{3^n +1}{2}\) 

Given, (1 – x + x2)n = a0 + a1x + a2x2 + ..... + a2n x2n ...(i) 

Putting x = 1 in the given equation (i), we have 

1 = a0 + a1 + a2 + a3 + ..... + a2n ...(ii) 

Putting x = –1 in the above given equation (i), we have

(1 – (–1) + 1)n = a0 – a1 + a2 – a3 + ..... + a2n 

⇒ 3n = a0 – a1 + a2 – a3 + ..... + a2n ...(iii)

Adding eqns (ii) and (iii), we have

1 + 3n = 2(a0 + a2 + a4 + ..... + a2n)

⇒ a0 + a2 + a4 + ..... + a2n

\(\frac{3^n +1}{2}\)

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