How to find the sum of $n(n+1)$, $(2n-1)$ and $(3n-2)$

Question

How can I find the next sums? $$\sum_{k=0}^n k(k+1)$$ $$\sum_{k=0}^n (2k-1)$$ $$\sum_{k=0}^n (3k-2)$$ How can I find their general formula? Maybe don't just lay it all out for me, but tell me how should I think? What should I try to do in order to find their formula?

Answer 1

You need to know the basic formulas $$ \sum _1^n k =n(n+1)/2$$ and $$\sum _1^n k^2=n(n+1)(2n+1)/6$$

Then you use the linearity of $\sum$ to write your problem in terms of the basic sums.

For example you get $$ \sum _1^n k(k+1)=\sum _1^n k^2+\sum _1^n k$$

Then you plug in the known values for the sums.

Answer 2

So, to formally prove it you'll use induction. As far as how to think about it, look for a pattern in the first several terms.

$$\sum k(k+1)=2+6+12+30+\dots$$

So, for instance, we can see right away that we can factor out a 2 and we have $2(1+3+6+15+\dots)$. Now can you come up with another pattern inside the parentheticals? (Notice that 3,6,15,... are all multiples of 3)

Answer 3

First, I would use these well-known formulas:

$$\sum_{k=0}^n k=\frac{n(n+1)}{2}$$ $$\sum_{k=0}^n k^2=\frac{n(n+1)(2n+1)}{6}$$

Then, you can do the rest in terms of these formulas by using linearity of summation. I will start the last one for you as an example:

$$\sum_{k=0}^n (3k-2)=\sum_{k=0}^n (3k)-\sum_{k=0}^n2=3\sum_{k=0}^nk-\sum_{k=0}^n2$$

I already told you $\sum_{k=0}^n k=\frac{n(n+1)}{2}$ and $\sum_{k=0}^n 2$ is just adding $2$ to itself $n+1$ times, so that's $2(n+1)$. Thus, now, just plug into the above and simplify to get the final answer. The other two sums can be solved very similarly.

Answer 4

Use $\sum k = \frac{k(k+1)}{2}$ and $\sum k^2 = \frac{k(k+1)(2k+1)}{6}$ to find the sums. If you need a formal proof of the sums just use that as the inductive hypothesis and prove it using induction.

Answer 5

Observe that

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

so that by telescoping,

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

Similarly,

$$n^2=\sum_{k=1}^n(2k-1)$$ (which by the way answers the second case), and

$$n=\sum_{k=1}^n 1.$$

Then by suitable linear combinations of $3k^2-3k+1,2k-1$ and $1$, you can answer for any quadratic polynomial.

Note that

$$\begin{pmatrix}3&-3&1\\0&2&-1\\0&0&1\end{pmatrix}^{-1}=\begin{pmatrix}\frac13&\frac12&\frac16\\0&\frac12&\frac12\\0&0&1\end{pmatrix}.$$