Answer to Question #267603 in Combinatorics | Number Theory for K12

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

Here $M_n=0.1+1.2+2.3+3.1+....+(n-1)n$

$n^{th}$ term is-

$t_n=(n-1)n =n^2-n$

So The Sum ,

$M_n=\sum t_n=\sum [n^2-n]=\sum n^2-\sum n$

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