Answer to Question #268577 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

$$\displaystyle\sum_{i=1}^ni=\dfrac{n(n+1)}{2}$$

$$\displaystyle\sum_{i=1}^ni^2=\dfrac{n(n+1)(2n+1)}{6}$$

$$\displaystyle\sum_{i=1}^n(i-1)i=\displaystyle\sum_{i=1}^n(i^2-i)=\displaystyle\sum_{i=1}^ni^2-\displaystyle\sum_{i=1}^ni$$

$$=\dfrac{n(n+1)(2n+1)}{6}-\dfrac{n(n+1)}{2}$$

$$=\dfrac{n(n+1)(2n+1-3)}{6}$$

$$=\dfrac{(n-1)n(n+1)}{3}$$

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

$$=\displaystyle\sum_{i=1}^n(i-1)i=\dfrac{(n-1)n(n+1)}{3}$$