Deriving formula for sum n(n+1)

Question

Can you please describe how to derive a formula for first n members of $$ S = 1\cdot 2 + 2\cdot 3 + 3\cdot 4 +\cdots +n(n+1)\mbox{?} $$ Thank you

Answer 1

Hint:

$$S = \sum\limits_{k=1}^{n}k(k+1)=\sum\limits_{k=1}^{n}(k^2+k) = \sum\limits_{k=1}^{n}k^2+\sum\limits_{k=1}^{n}k\text{.}$$ Do you know how to calculate the two sums $\displaystyle\sum\limits_{k=1}^{n}k^2$ and $\displaystyle\sum\limits_{k=1}^{n}k$?

Answer 2

Here's one method:

1. Guess that it's a polynomial.

2. If you want to figure out what degree it is, compute forward differences.

   That is, if $S(n)$ is the sum of the first $n$ terms, compute $D_1(n) = S(n) - S(n+1)$ for a few different $n$. Then go forward again, and compute $D_2(n) = D_1(n) - D_1(n+1)$ for a few $n$, and so on. Eventually, you might notice that the terms become constant. If this happens on $D_k$, then this suggests you have a degree $k$ polynomial.

3. If you don't want to figure it out, then it's a degree $3$ polynomial.

4. Find $4$ terms and fit a degree $3$ polynomial to those terms.

Answer 3

Hint: $$\sum _{i=k}^n\binom{i}{k}=\binom{n+1}{k+1}$$ Hope it helps.

Answer 4

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