^{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...

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 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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