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


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Matt L. | MIT/Harvard Grad for Academic Tutoring & Test PrepMIT/Harvard Grad for Academic Tutoring &...
5.0 5.0 (1077 lesson ratings) (1077)

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


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:

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


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.

Roman C. | Masters of Education Graduate with Mathematics ExpertiseMasters of Education Graduate with Mathe...
4.9 4.9 (346 lesson ratings) (346)

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 ...



Note that equality is approximate. I did this by completing the square:

x^2 + ax = x^2 + ax + (a/2)2 - (a/2)2 = (x + a/2)2 - (a/2)2.

In your problem we have a = 1.

Since 1/4 << (x+1/2)2, for large x, we can ignore 1/4 and take square roots to yield x+1/2. The error is very small, E << 1/2.