# Squared Summation(Intermediate Step)

I am studying Economics and are trying to get a firm grasp of summation rules and applications. Looking into the following relation,

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6} The following "trick" is given below, to understand the above. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$\sum_{k=1}^n[(k+1)^3-k^3]=n^3 + 3n^2+3n$

As I expand the left-handside of the equation for a given sequence $S=[1^2, 2^2, 3^3,..., n^3]$, the following leads to the right-hand side of the equation,

$(2^3 - 1^3) + (3^3 - 2^3) + (4^3 - 3^3) + .... + (n^3 - (n - 1)^3) + ((n+1)^3 - n^3) = (n + 1)^3 - 1^3 = n^3 + 3n^2 + 3n$

Understanding the intermediate steps, ie. the above expanding, Im struggling with the intuition so to speak. Which kind of mentality should I have had applied on the second equation, in order get to the right hand side, without expanding it?

Any help, with some mathematical explanation is highly appreciated.

Thank you!

You need to know also that $1+2+...+n=\frac{n(n+1)}{2}$ and use your work: $$n^3+3n^2+3n=3x+3\cdot\frac{n(n+1)}{2}+n,$$ which gives which you wish.
• Thank you Michael - I am aware of that. However, Im looking at the step before that. As in how the second equation, expands to $n^3 + 3n^2 + 3n$ skipping the hideous step-by-step calculation below. Jan 24, 2018 at 17:07
• You can use $(n+1)^3-1=(n+1-1)((n+1)^2+n+1+1)=n(n^2+3n+3).$ Also, $(k+1)^3-k^3=(k+1-k)((k+1)^2+(k+1)k+k^2)=3k^2+3k+1.$ Jan 24, 2018 at 17:10
I think this may be a question about how telescoping sums work. If so, then $$\sum_{k=1}^n\left(f(k+1)-f(k)\right)=\sum_{k=1}^nf(k+1)-\sum_{k=1}^nf(k)=\sum_{j=2}^{n+1}f(j)-\sum_{j=1}^nf(j)$$ Where e have made the substitution $k=j-1$ in the first sum and $j=k$ in the second. When $j-1=k=1$, $j=2$ and when $j-1=k=n$, $j=n+1$. Then $$\sum_{k=1}^n\left(f(k+1)-f(k)\right)=f(n+1)+\sum_{j=2}^nf(j)-f(1)-\sum_{j=2}^nf(j)=f(n+1)-f(1)$$ Where we have extracted the terms not common to both sums and canceled the common parts. In the instant case, $$\sum_{k=1}^n\left((k+1)^3-k^3\right)=(n+1)^3-1^3=n^3+3n^2+3n$$