Search
Ask a question
0

I can't understand how the infinity concept plays into this.

The value of the infinite product: 2^1/5 * 4^1/25 * 8^1/125 * 16^1/625 * ... can be expressed in simplest radical form as the nth root of a to the b power. Compute n+a+b.
 
I got up to the point where I know that the factors in the equation go in the formula 2n/5n, and I converted it to radical form, but I still don't know what n, a, and b are. Thanks!

2 Answers by Expert Tutors

Tutors, sign in to answer this question.
Richard P. | Fairfax County Tutor for HS Math and ScienceFairfax County Tutor for HS Math and Sci...
4.9 4.9 (587 lesson ratings) (587)
1
Check Marked as Best Answer
You are dealing with an infinite product.    Call the infinite product Q.    Convert the infinite product to an infinite sum by taking the log2 of both sides.     This results in:
 
log2(Q) =   ∑ n=1∞  n/5n  
 
This infinite sum can be found on Wikipedia.     The infinite sum is 1/(1-1/5)2 =  25/16.   So
 
log2(Q) =   25/16          Both side can be exponentiated with 2  to get
 
Q =   225/16     This is the 16th root of  ab   where a =2 and b = 25
 
So finally n =16,  a = 2 and b = 25.     Thus n+a+b =  43.
 
 
Phillip R. | Top Notch Math and Science Tutoring from Brown Univ GradTop Notch Math and Science Tutoring from...
0
the product is infinite so we have 2a + b + c + d.....
where a = 1/5, b = 2/25, c = 3/125, etc
 
in general the terms are n/5n
 
So the power of 2 is ∑ n/5n as n goes from 1 to ∞
 
This infinite series converges. The question is what is the value of this infinite sum.
I don't know if you are asking as a student in algebra 2 or calculus class.
 
If the sum = b/n and a = 2, then the infinite product = nth root of 2b
 
Perhaps a graphing calculator can give you the value for the infinite sum.