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Question

If f(n)=nr=11(r+r+1)(r1/4+(r+1)1/4) and g(n) is product of n terms of the G.P. 16,4,1, then which of the following is correct?


A
f(80) is a prime number
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B
f(9999)=10
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C
n=1(g(n))1/n=32
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D
limn(g(n))1/n2=132
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Solution

The correct options are
A f(80) is a prime number
C n=1(g(n))1/n=32
1(r+r+1)(r1/4+(r+1)1/4)=(r+1r)(r1/4+(r+1)1/4)=(r+1r)(r1/4(r+1)1/4)(r1/2(r+1)1/2)=(r+1)1/4r1/4
Therefore, 
f(n)=nr=1[(r+1)1/4r1/4]f(n)=(n+1)1/41
So, 
f(80)=(81)1/41=2
Which is a prime number.
f(9999)=(10000)1/41=9

Now,
g(n)=1641n termsg(n)=42+1+0+
We know that
2+1+0+n terms=n2[4+(n1)×(1)]=n(5n)2
Therefore, 
g(n)=2n(5n)n=1(g(n))1/n=n=12(5n)n=1(g(n))1/n=24+23+22+n=1(g(n))1/n=16112=32

limn(g(n))1/n2=limn2(5n)/n=12

Mathematics

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