Find the formula for the given sum of series

Question

Find the sum of the series:

$$\sum_{i=2}^{n}\binom{i}{2}= \,^{2}C_{2}+\cdots+\,^{n}C_{2}$$

I did try expanding it and see if I could simplify it further.I am unable to find a formula for it? Can anyone help me?

Answer 1

We have by the upper summation formula for binomial coefficients:

$$\sum_{i=2}^{n}\binom{i}{2}=\sum_{0\leq i \leq n}\binom{i}{2}=\binom{n+1}{3}=\frac{n(n-1)(n+1)}{6}$$

We can prove this identity fairly simply:

$$\sum_{0\leq i \leq n}\binom{i}{m}=\sum_{0 \leq m+i \leq n}\binom{m+i}{m}=\sum_{0 \leq i \leq n-m}\binom{m+i}{m}=\binom{m+(n-m)+1}{m+1}=\binom{n+1}{m+1}$$

You might want to check Chapter 5 of Knuth et al.'s _Concrete Mathematics for more binomial coefficient identities.

Answer 2

Here is a start

$$ \sum_{i=2}^{n}\binom{i}{2} = \sum_{i=2}^{n} \frac{i!}{2! (i-2)!}= \frac{1}{2}\sum_{i=2}^{n} i(i-1) = \frac{1}{2}\sum_{i=2}^{n} i^2 - \frac{1}{2}\sum_{i=2}^{n} i=\dots. $$

Can you finish it?

Note: You need the identities

$$ \sum_{i=1}^{n}i = \frac{n(n+1)}{2},\quad \sum_{i=1}^{n}i^2 = \frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}. $$

Answer 3

Write out the terms individually:

$$ \begin{align} \text{LHS} &= 1 + 3 + 6 + 10 + 15 + \cdots + \dfrac{n(n-1)}{2}\\ &= \sum_{i = 1}^{1}{i} + \sum_{i = 1}^{2}{i} + \cdots + \sum_{i = 1}^{n - 1}{i}\\ &= \sum_{i = 1}^{n - 1}{\sum_{j = 1}^{i}{i}}\\ &= \sum_{i = 1}^{n - 1}{\dfrac{i(i+1)}{2}}\\ &= \dfrac{(n + 1) n (n - 1)}{6} \end{align} $$