Binomial TheoremHard
Question
If recursion polynomials Pk(x) are defined as P1(x) = (x - 2)2, P2 (x) = ((x - 2)2 - 2)2
P3 (x) = ((x - 2)2 - 2)2 - 2)2 ..... (In general Pk (x) = (Pk-1 (x) - 2)2, then the constant term in Pk (x) is
P3 (x) = ((x - 2)2 - 2)2 - 2)2 ..... (In general Pk (x) = (Pk-1 (x) - 2)2, then the constant term in Pk (x) is
Options
A.4
B.2
C.16
D.a perfect square
Solution
Constant term in P1(x) is 4
If the constant term in Pk(x) is also 4, then
Pk(x) = 4 + a1x + a2x2 + .........
and Pk+1(x) = (Pk(x) -2)2 = (a1x + a2x2 + ....+ 2)2
If the constant term in Pk(x) is also 4, then
Pk(x) = 4 + a1x + a2x2 + .........
and Pk+1(x) = (Pk(x) -2)2 = (a1x + a2x2 + ....+ 2)2
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