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Question

let x=(1000000000000000000001)²⁰ then, under root x²+x is equal to what?

 

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Answer 1

Roman is right that the answer is very close to

(1000000000000000000001)²⁰ + 1/2 = (10²¹+1)²⁰ + 1/2

In particular, he's right that the error is very small, much less than 1/2. But you can be even more precise using the Arithmetic Mean-Geometric Mean-Harmonic Mean (AM-GM-HM) Inequality. That's because

sqrt(x²+x)=sqrt[x(x+1)]

is just the geometric mean of x and x+1. The AM-GM-HM inequality says that the arithmetic mean of two different numbers is greater than the geometric mean which in turn is greater than harmonic mean. In other words, the geometric mean is between the arithmetic mean and the harmonic mean. If you don't know what these means are, you can read about them here:

http://www.artofproblemsolving.com/Wiki/index.php/RMS-AM-GM-HM

In the problem you gave, the arithmetic mean of x and x+1 is just x + 1/2, and after a little algebra you'll see that the harmonic mean is

[x(x+1)]/(x+1/2)

That means the difference between the arithmetic mean and the harmonic mean is

AM - HM = (x + 1/2) - [x(x+1)]/(x+1/2) = [(x+1/2)^2-x(x+1)]/(x+1/2)=(1/4)/(x+1/2)=1/(4x+2)

For the gigantic number x you gave, namely x=(10²¹+1)²⁰, this gap between the arithmetic mean and the harmonic mean is astronomically small. Since the geometric mean, which is what we want, must be between the arithmetic mean and the harmonic mean, the estimate of the geometric mean as x + 1/2 is extremely, extremely close.

Answer 2

Note √(x²+x) ≈ x + 1/2 (the error in the estimate approaches 0 as x approaches infinity) because x²+x = (x+1/2)² - 1/4. So the answer is approximately (1000000000000000000001)²⁰ + 1/2

We can expand this as a decimal using binomial coefficients C(20,n) for n=1,...,20 and get

1000000000000000000020000000000000000000190000000000000000001140 ...

000000000000000004845000000000000000015504000000000000000038760 ...

000000000000000077520000000000000000125970000000000000000167960 ...

000000000000000184756000000000000000167960000000000000000125970 ...

000000000000000077520000000000000000038760000000000000000015504 ...

000000000000000004845000000000000000001140000000000000000000190 ...

000000000000000000020000000000000000000001.5