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Question

The sum of the co-efficients of all odd degree terms in the expansion of (x+x31)5+(xx31)5,(x>1) is

A
1
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B
2
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C
1
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D
0
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Solution

The correct option is D 2
(x+x31)5+(xx31)5

(a+b)5=a5+5a4b+10a3b2+10a2b3+5ab4+b5(i)

(ab)5=a55a4b+10a3b210a2b3+5ab4b5(ii)

(a+b)5+(ab)5=2[a5+10a3b2+5ab4]

=2[x5+10x3(x31)+5x(x31)4]

=2[x5+10x610x3+5x(x62x3+1)]

=2x5+20x620x3+10x720x4+10x

Here all the co-efficient of the above equation are odd terms coefficients of even terms are cancelled out.
So, sum of coefficients 2+2020+1020+10=2

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