Estimate the sum to the nearest whole number

Dec 2016
295
161
Earth
Estimate the sum to the nearest whole number without the use of a calculator/computer.


\(\displaystyle \displaystyle\sum_{n = 625}^{9999} \sqrt[4]{n + 0.5}\)
 
  • Like
Reactions: 1 person

romsek

MHF Helper
Nov 2013
6,662
2,999
California
amazing

eagerly awaiting the simple explanation
 

Debsta

MHF Helper
Oct 2009
1,297
592
Brisbane
Estimate the sum to the nearest whole number without the use of a calculator/computer.


\(\displaystyle \displaystyle\sum_{n = 625}^{9999} \sqrt[4]{n + 0.5}\)
First thoughts. Here's one estimate:

If \(\displaystyle n=5\) then \(\displaystyle \sqrt[4]{n+0.5} \approx \sqrt[4]{625} = 5 \)

If \(\displaystyle n=9999\) then \(\displaystyle \sqrt[4]{n+0.5} \approx \sqrt[4]{10000} = 10 \)


So the first number in the sum is approx 5 and the last number is approx 10. As n increases, the value of \(\displaystyle \sqrt[4]{n+0.5}\) 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!
 
Last edited:
  • Like
Reactions: 1 person

Debsta

MHF Helper
Oct 2009
1,297
592
Brisbane
A (much) better estimate:


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

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

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

\(\displaystyle = 77 500 \)
 
Last edited:
  • Like
Reactions: 3 people