Squared Summation(Intermediate Step)

Question

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!

Answer 1

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.

Answer 2

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, $$\sum_{k=1}^n\left((k+1)^3-k^3\right)=(n+1)^3-1^3=n^3+3n^2+3n$$