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Question

The sum of the coefficients of even power of x in the expansion of (1+x+x2+x3)5 is

A
64
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B
512
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C
128
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D
256
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Solution

The correct option is B 512
(1+x+x2+x3)5=a0+a1x+a2x2+a3x3++a15x15

Putting x=1 and x=1 alternatively, we get
a0+a1+a2+a3++a15=45(1)a0a1+a2a3+a15=0(2)
Adding equation (1) and (2), we have 2(a0+a2+a4++a14)=45a0+a2+a4++a14=29a0+a2+a4++a14=512

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