Answer to Question #269276 in Algebra for Random

Question

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 rst n squares:

Qn = 1 .1 + 2 . 2 + 3 .3 + ::: + n .n = n3/3 + n2/2 + 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+4+.......+n =$\frac{( n²+n)}{2}$

Q_n = 1.1 + 2.2 +3.3+......+n.n = $\frac{n³}{3}+\frac{n²}{2}+\frac{n}{6}$

So Q_n - Sn =$[\frac{n³}{3}+\frac{n²}{2}+\frac{n}{6}]-\frac{( n²+n)}{2}$

=> {1.1 + 2.2 +3.3+......+n.n } - {1+2+3+.......+n} = $\frac{n³}{3}+\frac{n²}{2}+\frac{n}{6}-\frac{n²}{2}-\frac{n}{2}$

Rearranging the left hand side

(1.1-1)+(2.2-2)+(3.3-3)+....+(n.n-n) = $\frac{n³}{3}-\frac{n}{3}$

=> 1(1-1)+2(2-1)+3(3-1)+.....+n(n-1) = $\frac{n³}{3}-\frac{n}{3}$

=> 1.0+2.1+3.2+4.3+....+n(n-1) = $\frac{n(n²-1)}{3}$

=> 0.1+1.2+2.3+3.4+....+(n-1).n = $\frac{n(n²-1)}{3}$

So M_n = $\frac{n(n²-1)}{3}$