If n is an integer greater than 1, what is the value of 10(root10)

If you look at all of those questions, you'll notice not one of them is official. The GMAT never tests infinite sums and products like the ones you see in these questions, because to answer questions like this, first you need to establish that these infinite expressions "converge" (they don't become infinite or undefined) and to do that, you need calculus. Calculus is not tested on the GMAT.

So these questions are all out of scope, and I don't see how the technique one uses to solve them could be relevant to any actual GMAT question, but if we need to solve, we can do as follows. Let x equal the expression:

\(\\
x = 10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{...}}}}\\
\)

Now notice that after the first "\(\sqrt[n]{ }\)", what we see is identical to x. So we can replace it with "x", and we have

\(x = 10*\sqrt[n]{x}\)

Now we just need to solve this equation: x/10 = x^(1/n), so x^n / 10^n = x, and x^(n-1) = 10^n, so x = 10^(n/(n-1))

With these answer choices, it's even faster to notice that we're multiplying 10 here by something greater than 1. So the answer must be greater than 10, and so must be D or E. But the larger n is, the smaller the roots all get, so the smaller the eventual answer should be. So E can't be right, because E gets bigger if you make n bigger, and only D is possible.

Solution

• Given n is an integer greater than 1

o This means, \(\frac{1}{n} < 1\)
• Now, \(10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{….,}}}} = 10*10^{\frac{1}{n}}*10^{\frac{1}{n}*\frac{1}{n}}*10^{\frac{1}{n}*\frac{1}{n}*\frac{1}{n}}* ……,\)

o \(⟹ 10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{….,}}}} = 10^{1+\frac{1}{n} + \frac{1}{n^2} +\frac{1}{n^3}…..}\)
• We can observe that \({\frac{1}{n} + \frac{1}{n^2} +\frac{1}{n^3}…..}\) is an infinite G.P.,

o \(⟹ 10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{….,}}}} = 10^{1+\frac{1}{n} + \frac{1}{n^2} +\frac{1}{n^3}…..}\) = \( 10^{\frac{1}{(1-\frac{1}{n})}}\) = \(10^{\frac{n}{n-1}}\)

Thus, the correct answer is Option D.

Solution:

Letting x = the expression, we can create the equation:

x = 10 * n^√x

x = 10x^(1/n)

x^(1 - 1/n) = 10

x^[(n - 1)/n] = 10

x = 10^[n/(n - 1)]

Answer: D

Official Solution:

If \(n\) is an integer greater than 1, what is the value of \(10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{...}}}}\), where the given expression extends to an infinite number of roots?

A. \(10\)
B. \(10^{\frac{1}{n}}\)
C. \(10^{\frac{n-1}{n}}\)
D. \(10^{\frac{n}{n-1}}\)
E. \(10^{n}\)


Let \(x=10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{...}}}}\)

Now, re-write above as \(x=10*\sqrt[n]{(10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{...})}}}\).

Since the expression extends to an infinite number of roots, then the expression in brackets would also equal to \(x\). Thus we can replace the expression in brackets with \(x\) and rewrite the expression as: \(x=10*\sqrt[n]{x}\)

Take above to the \(n^{th}\) power:

Take \(n-1^{th}\) root:


Answer: D

Asked: If n is an integer greater than 1, what is the value of \(10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{...}}}}\), where the given expression extends to an infinite number of roots?

Let x= \(10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{...}}}}\)

\(x = 10*\sqrt[n]{x}\)

\(x^n = 10^n*x\)
\(x^{n-1} = 10^n\)
\(x = 10^{\frac{n}{n-1}}\)

IMO D