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Question

Let (x3+x5)(2x6+3x41)20=a0+a1x+a2x2........+a125x125. Then
a0a23+a45a67.......+a124125=

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Solution

Given,
(x3+x5)(2x6+3x41)20=a0+a1x+a2x2........+a125x125.
or, x(x2+x4)(2x6+3x41)20=a0+a1x+a2x2........+a125x125.
Now integrating both sides between the limits 1 and 1 we get,
11x(x2+x4)(2x6+3x41)20dx=11(a0+a1x+a2x2........+a125x125)dx
or, 0=2(a0a23+a45a67.......+a124125) [ Using 11f(x)dx=0 for f(x) being odd and 11f(x)dx=210f(x)dx for f(x) being even]
or, (a0a23+a45a67.......+a124125)=0.

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