Answer to Question #269311 in Calculus for Unknown356697

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

$M_n=0\cdot1+1\cdot2+2\cdot3+3\cdot4....(n-1)n$

$0\cdot1=0$

$0\cdot1+1\cdot2=0+2=2=\frac{1}{3}*(2-1)*(2+1)*2$

$0\cdot1+1\cdot2+2\cdot3=0+2+6=8=\frac{1}{3}(3-1)*(3+1)*3$

$0\cdot1+1\cdot2+2\cdot3+3\cdot4=0+2+6+12=20=\frac{1}{3}(4-1)*(4+1)*4$

From the above relations of $M_n=0\cdot1+1\cdot2+2\cdot3+3\cdot4....(n-1)n$

We obtain $M_n=\frac{1}{3}(n-1)(n+1)n$

$\therefore M_n=\frac{1}{3}n(n^2-1)$