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Question
If (1 – x + x2)n = a0 + a1 x + a2 x2 + ... + a2n x2n , then a0 + a2 + a4 + ... + a2n equals ______.
Options
`(3^"n" + 1)/2`
`(3^"n" - 1)/2`
`(1 - 3^"n")/2`
`3^"n" + 1/2`
Solution
If (1 – x + x2)n = a0 + a1 x + a2 x2 + ... + a2n x2n , then a0 + a2 + a4 + ... + a2n equals `(3^"n" + 1)/2`.
Explanation:
Putting x = 1 and –1 in
(1 – x + x2)n = a0 + a1 x + a2 x2 + ... + a2n x2n
We get 1 = a0 + a1 + a2 + a3 + ... + a2n ......(1)
And 3n = a0 – a1 + a2 – a3 + ... + a2n ......(2)
Adding (1) and (2), we get
3n + 1 = 2(a0 + a2 + a4 + ... + a2n)
Therefore a0 + a2 + a4 + ... + a2n = `(3^"n" + 1)/2`
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