^{th}root of a to the b power. Compute n+a+b.

^{n/5n}, and I converted it to radical form, but I still don't know what n, a, and b are. Thanks!

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 n^{th} 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 2^{n/5n}, and I converted it to radical form, but I still don't know what n, a, and b are. Thanks!

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Richard P. | Fairfax County Tutor for HS Math and ScienceFairfax County Tutor for HS Math and Sci...

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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 log_{2} of both sides. This results in:

log_{2}(Q) = ∑ _{n=1}^{∞ } n/5^{n}

This infinite sum can be found on Wikipedia. The infinite sum is 1/(1-1/5)^{2} = 25/16. So

log_{2}(Q) = 25/16 Both side can be exponentiated with 2 to get

Q = 2^{25/16 } This is the 16th root of a^{b} where a =2 and b = 25

So finally n =16, a = 2 and b = 25. Thus n+a+b = 43.

the product is infinite so we have 2^{a + b + c + d.....}

where a = 1/5, b = 2/25, c = 3/125, etc

in general the terms are n/5^{n}

So the power of 2 is ∑ n/5^{n} 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 = n^{th} root of 2^{b}

Perhaps a graphing calculator can give you the value for the infinite sum.

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