Answer to Question #268576 in Calculus 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

The sum Mn can be written as follows:

$$M_n = \sum_{k=1}^n(k-1)k$$

Expanding the expression under the sum sign, obtain:

$$M_n=\sum_{k=1}^nk^2-k = \sum_{k=1}^nk^2-\sum_{k=1}^nk$$

The first term is the sum of squares and the second term is the sum of natural numbers. Thus, obtain:

$$M_n= Q_n-S_n=\dfrac{n(n+1)(2n+1)}{6} - \dfrac{n(n+1)}{2}=\\ =\dfrac{n(n+1)(2n+1)-3n(n+1)}{6} = \dfrac{n(n+1)(2n+1-3)}{6}=\\ = \dfrac{n(n+1)(2n-2)}{6} = \dfrac{n(n+1)(n-1)}{3}=\\ = \dfrac{n(n^2-1)}{3}$$

Answer. $M_n = \dfrac{n(n^2-1)}{3}$.