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Question

# If f(n)=n∑r=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=321(√r+√r+1)(r1/4+(r+1)1/4)=(√r+1−√r)(r1/4+(r+1)1/4)=(√r+1−√r)(r1/4−(r+1)1/4)(r1/2−(r+1)1/2)=(r+1)1/4−r1/4 Therefore, f(n)=n∑r=1[(r+1)1/4−r1/4]⇒f(n)=(n+1)1/4−1 So, f(80)=(81)1/4−1=2 Which is a prime number. f(9999)=(10000)1/4−1=9 Now, g(n)=16⋅4⋅1⋯n terms⇒g(n)=42+1+0+⋯ We know that 2+1+0+⋯n terms=n2[4+(n−1)×(−1)]=n(5−n)2 Therefore, g(n)=2n(5−n)⇒∞∑n=1(g(n))1/n=∞∑n=12(5−n)⇒∞∑n=1(g(n))1/n=24+23+22+⋯∴∞∑n=1(g(n))1/n=161−12=32 limn→∞(g(n))1/n2=limn→∞2(5−n)/n=12

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