How to find $\sum n^3$ if $\sum n^2$ is given

Problem :

Find $\sum_{j=1}^n j^3$ if $\sum_{j=1}^n j^2 =2870$

Can we use the following method :

$\sum_{j=1}^n j^2 = \frac{n(n+1)(2n+1)}{6}$ = 2870..

( As sum of the square of first n natural number is $\frac{n(n+1)(2n+1)}{6}$)

But how do we proceed from here to get the result?

• You could factor 2870. Then identify the factors with those in $\frac{n(n+1)(2n+1)}{6}$ to find $n$. Jul 31, 2013 at 12:58
• Do you know the formula for $\sum n^3$ ? Jul 31, 2013 at 13:03
• yeah it is $[\frac{n(n+1)}{2}]^2$ @ Thomas Jul 31, 2013 at 13:04
• See here. Jul 31, 2013 at 13:08
• I've added (summation) tag, which is intended specifically for finite sums (among other things). Jul 31, 2013 at 13:47

Note that $n(n+1)(2n+1)=2n(n+1)(n+1/2)$. Letting $m=n+1/2$, we get:

$$2780\cdot 3 = (m-1/2)(m+1/2)m = m^3-m/4$$ So we want $m$ to be just over $\sqrt[3]{3\cdot 2780}$. That should give you a candidate for $n$. You should check that this $n$ works.

Note: The $m$ trick is a bit of magic, but we know that $n(n+1)(2n+1)$ grows "like" $2n^3$, so even if you don't find the $m$ trick, you can guess it will be near $\sqrt[3]{3\cdot 2780}$. The $m$ trick makes it clear that if such $n$ exists, there has to be only a single value.

First we factor $2870$: $$2870=(20\times 21\times 41)/6$$ We have $$(k+1)^4-k^4= 4k^3+6k^2+4k+1$$ so $$\sum_{k=1}^{n=20}(k+1)^4-k^4=(21)^4-1=4\underbrace{\sum_{k=1}^{n=20} k^3}_{=S}+6\underbrace{\sum_{k=1}^{n=20}k^2}_{=2870}+4\underbrace{\sum_{k=1}^{n=20}k}_{=20\times21/2}+\underbrace{\sum_{k=1}^{n=20}1}_{=20}$$ and then we find $S$.

• One more "for the road"! Apr 26, 2014 at 12:42

One more way: 1) use perturbation method (see Concrete Mathematicxs of and 2) solve the equation you have for $n$.

The equation $2n^3+2n^2+n-6 \cdot 2870=0$ has only 1 real root: $n=20$. Now use perturbation method, denote $S_n=\sum_{k=1}^{n}k^4$ and $S^{\ast}_n=\sum_{k=1}^{n}k^3$ $$S_n +(n+1)^4=\sum_{k=1}^{n}(k+1)^4+1=S_n+4 \sum_{k=1}^{n}k^3+6 \sum_{k=1}^{n}k^2+4\sum_{k=1}^{n}k+n+1$$ Here $S_n$ cancel out, you have thee value for $\sum_{k=1}^{n}k^2$, from which you can also find the value for $\sum_{k=1}^nk$ (it is 210), so you can easily find the value for $S^{\ast}_n$.

• Why to use perturbation method if $n$ is known? Aug 3, 2018 at 9:27