Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a really confusing question from an investigation. It states-

Find the value of:


How would I go about answering this??

share|cite|improve this question

closed as off-topic by Rory Daulton, Daniel W. Farlow, Jeel Shah, N. F. Taussig, ᴡᴏʀᴅs Feb 27 at 23:34

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Rory Daulton, Daniel W. Farlow, Jeel Shah, N. F. Taussig
If this question can be reworded to fit the rules in the help center, please edit the question.

Welcome to MathSE! You are more likely to get a good answer to your question if you follow a few guidelines. In particular, what have you tried so far, and just where are you stuck? This is not a homework-answering site: we want to see that you have put significant work into the problem. For example, have you tried to come up with a formula for adding the first $n$ cubic numbers? – Rory Daulton Feb 27 at 16:54

$$\sum_{r=1}^n r^3=\frac {n^2(n+1)^2}{2^2}=\left(\sum_{r=1}^nr\right)^2\\ \sqrt{\sum_{r=1}^{100}r^3}=\sum_{r=1}^{100}r=\binom{101}2=5050$$ i.e. $$\sqrt{1^3+2^3+3^3+\cdots+100^3}=1+2+3+\cdots+100=5050$$

share|cite|improve this answer

Hint: show by induction that $$\sum_{i=1}^n i^3 = \frac{n^2(n+1)^2}{4}.$$

Note that the right-hand side is just $$\left(\sum_{i=1}^n i\right)^2 = \left(\frac{n(n+1)}{2}\right)^2.$$

share|cite|improve this answer


Prove by induction that $$1^3+2^3+3^3+\ldots + n^3 = \left[\frac{n(n+1)}{2}\right]^2$$ and use that result in your question.

share|cite|improve this answer

$$\sum_{i=1}^n i^3 = \frac{n^2(n+1)^{2}}{4}$$

Can you proceed from here?

share|cite|improve this answer

And if you don't know the formula and don't need it exactly,

$\sum_{k=1}^{100} k^3 \approx \int_0^{100} x^3 dx =\frac{100^4}{4} $ so the result is $\sqrt{\frac{100^4}{4}} =\frac{100^2}{2} =5000 $.

If you add in the usual correction of $\frac12 f(n)$, the result is $\sqrt{\frac{100^4}{4}+\frac12 100^3} =\frac{100^2}{2}\sqrt{1+\frac{2}{100}} \approx \frac{100^2}{2}(1+\frac{1}{100}) =5050 $.


share|cite|improve this answer

There is a nice visual proof (of concept) of the Nicomachus's theorem: the sum of first cubes is a squared triangular number:

Wikipedia: Nichomacus theorem

So the square root of the total volume of the cubes above is the side of the square below, hence the sum of the first integers.

In this example, you have $225$ "unit cubes", whose root is $15$, i. e. the length of the side of the square below.

It takes some more time to draw $100$ increasing cubes, but the proof works along the same lines.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.