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Question

If limx2nr=1xrnr=12rx2=f(n), then which of the following is/are CORRECT?

A
f(2020)=1999(2)2020+1
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B
f(2020)=505(2)2022+1
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C
f(6)=321
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D
f(10)=9217
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Solution

The correct option is D f(10)=9217
Given : limx2nr=1xrnr=12rx2

Now,
f(n)=limx2(x2)x2+(x222)x2++(xn2n)x2f(n)=1+22+322++n2n1(i)2f(n)=2+222++(n1)2n+n2n(ii)
Subtracting (i) and (ii)
f(n)=(1+2+22+2n1)n2nf(n)=1(2n1)21n2nf(n)=2n1n2nf(n)=(n1)2n+1

Hence,
f(2020)=2019(2)2020+1f(6)=526+1=321f(10)=9210+1=9217

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