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Question

# Consider (1+x)2n+(1+2x+x2)n=2n∑r=0arxr,n∈N. If 2n∑r=0ar=f(n), then

A
n=11f(n)=16
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B
n=11f(n)=38
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C
largest value of p for which f(5) is divisible by 2p is 11
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D
largest value of p for which f(5) is divisible by 2p is 9
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Solution

## The correct options are A ∞∑n=11f(n)=16 C largest value of p for which f(5) is divisible by 2p is 11(1+x)2n+(1+2x+x2)n=2n∑r=0arxr ⇒(1+x)2n+(1+x)2n=2n∑r=0arxr ⇒2(1+x)2n=2n∑r=0arxr Put x=1, we get f(n)=2n∑r=0ar=22n+1 ⋯(1) Now, ∞∑n=11f(n)=123+125+127+⋯ =1231−14 =16 From (1), f(5)=211 Largest value of p for which f(5) is divisible by 2p is 11.

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