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
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
f(9999)=10
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
∞∑n=1(g(n))1/n=32
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
limn→∞(g(n))1/n2=132
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct options are Af(80) is a prime number C∞∑n=1(g(n))1/n=32 1(√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⋯nterms⇒g(n)=42+1+0+⋯ We know that 2+1+0+⋯nterms=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