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Question

The greatest term (numerically) in the expansion of (35x)11 when x=15 is

A
55×39
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B
46×39
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C
55×36
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D
None of these
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Solution

The correct option is A 55×39
We have,
(35x)11=311(15x3)11=311(113)11(x=15)
m=|x|(n+1)(|x|+1)=(13)(11+1)(13+1)=3
The greatest term in the expansion are T3 and T4
Greatest term ( when r=2) =311|T2+1|
=31111C2(13)2=31111.101.2×19=55×39
and greatest term ( when r3) =311|T2+1|
=31111C3(13)3=31111.10.91.2.3×127=55×39
From above we see that the values of both greatest term are equal

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