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# 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

Richard P. | Fairfax County Tutor for HS Math and ScienceFairfax County Tutor for HS Math and Sci...
4.9 4.9 (628 lesson ratings) (628)
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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...
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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.