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Question

In the expansion of (x31x2)n, nN, if the sum of the coefficients of x5 and x10 is 0, then n is

A
25
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B
20
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C
15
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D
10
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Solution

The correct option is C 15
(x31x2)n
General term Tr+1=n!r!(nr)!(1)nrx5r2n
If 5r2n=5, then 5r=2n+5 or r=2n5+1

If 5r2n=10, then 5r=2n+10 or r=2n5+2
x5 and x10 terms occur if n=5k

Given, the sum of coefficients x5 and x10 is zero.
(5k)!(2k+1)!(3k1)!(5k)!(2k+2)!(3k2)!=0
or, 13k112k+2=0
or, k=3n=15

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