Answer to Question #268579 in Combinatorics | Number Theory for K11

Question

The formula for calculating the sum of all natural integers from 1 to n is well-known:

Sn = 1 + 2 + 3 + ... + n =

n

2 + n

2

Similary, we know about the formula for calculating the sum of the first n squares:

Qn = 1 · 1 + 2 · 2 + 3 · 3 + ... + n · n =

n

3

3

+

n

2

2

+

n

6

Now, we reduce one of the two multipliers of each product by one to get the following sum:

Mn = 0 · 1 + 1 · 2 + 2 · 3 + 3 · 4 + ... + (n − 1) · n

Find an explicit formula for calculating the sum Mn.

Answer 1

$S_n = 1+2+3+...+n = \frac{n^2+n}{2} \\ Q_n = 1 \cdot 1+2 \cdot 2+3 \cdot 3+...+n \cdot n = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} \\ M_n = 0 \cdot 1 + 1 \cdot 2 +2 \cdot 3 +...+(n-1)n \\ = (1-1) \cdot 1 + (2-1) \cdot 2 + (3-1) \cdot 3 + ...+(n-1) \cdot n \\ = (1 \cdot 1 -1 \cdot 1) + (2 \cdot 2 -1 \cdot 2) + (3 \cdot 3- 1 \cdot 3) + … + ( n \cdot n -1 \cdot n) \\ = 1 \cdot 1 + 2 \cdot 2 + 3 \cdot 3 +...+ n \cdot n -(1 \cdot 1 + 1 \cdot 2 + 1 \cdot 3 +...+ 1 \cdot n) \\ = Q_n -S_n \\ = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} -(\frac{n^2+n}{2}) \\ = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} -\frac{n^2}{2} -\frac{n}{2} \\ = \frac{n^3}{3} + \frac{n-3n}{6} \\ = \frac{n^3}{3} + \frac{-2n}{6} \\ = \frac{n^3}{3} - \frac{n}{3} \\ M_n = \frac{n^3}{3} -\frac{n}{3}$