Thread: Estimate the sum to the nearest whole number

1. Estimate the sum to the nearest whole number

Estimate the sum to the nearest whole number without the use of a calculator/computer.

$\displaystyle \displaystyle\sum_{n = 625}^{9999} \sqrt{n + 0.5}$

2. Re: Estimate the sum to the nearest whole number

amazing

eagerly awaiting the simple explanation

3. Re: Estimate the sum to the nearest whole number Originally Posted by greg1313 Estimate the sum to the nearest whole number without the use of a calculator/computer.

$\displaystyle \displaystyle\sum_{n = 625}^{9999} \sqrt{n + 0.5}$
First thoughts. Here's one estimate:

If $\displaystyle n=5$ then $\displaystyle \sqrt{n+0.5} \approx \sqrt{625} = 5$

If $\displaystyle n=9999$ then $\displaystyle \sqrt{n+0.5} \approx \sqrt{10000} = 10$

So the first number in the sum is approx 5 and the last number is approx 10. As n increases, the value of $\displaystyle \sqrt{n+0.5}$ also increases.

There are 9999 - 625 +1 = 9375 terms in the sum.

Let's assume it is an AP (it isn't, but this will give us an approximation of the sum since the range of values 5 to 10 is quite small and there are a lot of terms).

So, using the sum of an AP formula, Sum $\displaystyle \approx \frac{9375}{2}(5+10) \approx70313$

(This gives the same result as if we say each term is approx 7.5 (ie halfway between 5 and 10)).

So an estimate is 70313.

Of course there would be other ways and this probably isn't the best!

4. Re: Estimate the sum to the nearest whole number

A (much) better estimate:

$\displaystyle \int_{625}^{9999} (n+0.5)^\frac{1}{4} dn$

$\displaystyle \approx \frac{4}{5} (10000^ \frac{5}{4} - 625^\frac{5}{4})$

$\displaystyle = \frac{4}{5} (10^5 - 5^5)$

$\displaystyle = 77 500$