Answer to Question #269569 in Algebra for Tahmina

Question

Problem A.3

The formula for calculating the sum of all natural integers from 1 to n is well-known: Sn =1+2+3+...+n= n2 +n

 2

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

n3 n2 n Qn =1·1+2·2+3·3+...+n·n= 3 + 2 + 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

Given that sum of all natural numbers 1 to n is:

$S_{n} = 1 + 2 + 3+ ..... + n = \frac{n\\^2 + n }{2} ------- (1)$

Sum of first n natural numbers square is:

$S_{n} = 1.1 + 2.2 + 3.3 + ..... + n.n \\ \frac{n\\^3}{3} + \frac{n\\^2}{2}+ \frac{n}{6}= .................. (2)$

Substract equation (1) from equation (2)

$=(1.1 + 2.2 + 3.3 + ..... + n.n) - (1 + 2 + 3+ ..... + n)$

$=(1.1 - 1+ 2.2 -2 + 3.3 -3 + ..... + n.n - n)$

$= \frac{n\\^3}{3} + \frac{n\\^2}{2}+ \frac{n}{6} - \frac{n\\^2 + n }{2}$

$=0+ 2(2 -1) + 3(3 -1) + ..... + n(n - 1)$

$=0+ 1. 2 + 2.3 + ..... + n(n - 1)$

$=\frac{ 2n\\^3 - 2n}{6}$

$=\frac{ n\\^3 - n}{3}$

Th explicit formula for calculating the sum Mn is $=\frac{ n\\^3 }{3} - \frac{n}{3}$