Binomial TheoremHard
Question
If (1 + x)10 = a0 + a1x + a2x2 +......+ a10x10, then value of (a0 - a2 + a4 - a6 + a8 - a10)2 + (a1 - a3 + a5 - a7 + a9)2 is
Options
A.210
B.2
C.220
D.None of these
Solution
(1 + x)10 = a0 + a1 + a2x2 +...... + a10x10
Put x = i,
(1 + i)10 = a0 - a2 + a4 + ....+ a10 + i (a1 - a3 + .....+ a9)
a0 - a2 + a4 + ....+ a10 = real part of (1 + i)10 = 25cos10π/4
a1 - a3 + ....... = imaginary part of (1 + i)10 = 25 sin10π/4 ....(2)
(1)2 + (2)2 = 210
Put x = i,
(1 + i)10 = a0 - a2 + a4 + ....+ a10 + i (a1 - a3 + .....+ a9)
a0 - a2 + a4 + ....+ a10 = real part of (1 + i)10 = 25cos10π/4
a1 - a3 + ....... = imaginary part of (1 + i)10 = 25 sin10π/4 ....(2)
(1)2 + (2)2 = 210
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