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$$S=\sqrt{2000^2-1^2}+\sqrt{2000^2-2^2}+\sqrt{2000^2-3^2} + \ldots + \sqrt{2000^2 - 2000^2}$$

For $S$ as defined above, find $\dfrac{2}{2000^2}\times(2000+2\times S)$ to $2$ decimal places.

Give solution with explanation and do not use calculator.

I am a student of class 10 and I reside in India. This is a question that I came up with. I asked this question to many of my seniors and all they said that it could be done only using Integration. But there is a different way also that we can solve it without using calculus. Since there is not enough space to write the proof as it is lengthy I would summarize it.

Please bear with me as I am not an experienced in writing proofs.

  • First, We take a semi-circle and then I drew chords parallel to the diameter.
  • Second, From any 2 consecutive parallel chords we can make a trapezium with really small height and then using Pythagoras Theorem we could find the length of each chord.
  • Third, We would then find the sum of the areas of all the trapeziums so that we can approximate the area of the semicircle.

After we have done it all we would reach this equation,

$\tfrac{\pi *r^2}{2}=c*(r+2\sum_{n=1}^{r-1}\sqrt{r^2-c^2*n^2})$

Where, r is the radius, and c is the height of the trapeziums.

Now, by observation we would find the following inequality for the height i.e. 'c':

$1

If we take larger values of the radii such as 2000 we would find that the value of 'c' gets closer and closer to 1.

Now, we can just put c=1 for the value of radius is large i.e. 2000.

So, we would get,

${\pi}=(r+2\sum_{n=1}^{r-1}\sqrt{r^2-n^2})*\frac{2}{r^2}$

and now we can find the answer of the question.

If you want me to write a more elaborate proof then please say so but it would take some time.

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    Did you try something?People most likely won't bother to look a question which shows no attempts to answer the question.Also consider not using imperative statements. – 2017-02-18
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    Back at you: Give context, with explanation. – 2017-02-19

1 Answers 1

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Using trapezoidal rule

\begin{align*} f(x) &= \sqrt{1-x^2} \\ \int_{0}^{1} f(x) \, dx & \approx \frac{1}{n} \sum_{k=0}^{n} f\left(\frac{k}{n} \right)-\frac{f(0)+f(1)}{2n} \\ \frac{\pi}{4} & \approx \frac{1}{2000}\sum_{k=0}^{2000} \sqrt{1-\frac{k^2}{2000^2}}- \frac{1}{2 \times 2000} \\ &= \frac{1}{2000^2}\sum_{k=1}^{2000} \sqrt{2000^2-k^2}+ \frac{1}{2 \times 2000} \\ &= \frac{S}{2000^2}+\frac{1}{2\times 2000} \\ \frac{4S}{2000^2}+\frac{2}{2000} &\approx \pi \end{align*}

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    nice solution. Also consider looking mine. – 2017-02-19