Prove by math induction that <span title="n.1+(n-1).2+(n-2).3+....+2(n-1)+1.n=n.(n+1).(n+2)/6?">n.1+(n-1).2+(n-2).3+....+...</span>

Question

Prove by math induction that n.1+(n-1).2+(n-2).3+....+...

Answer 1

general term is a[j] = (n+1-j)*j; s(n) =Σa[j], where j=1 until n;

♠check n=2; s(2) =(2+1-1)*1 +(2+1-2)*2 = 4;

and 2*(2+1)*(2+2)/6 = 4; good!

♠check n=3; s(3) =(3+1-1)*1 +(3+1-2)*2 +(3+1-3)*3 = 10;

and 3*(3+1)*(3+2)/6 =10; good!

♣ s(n+1) =Σ ((n+1) +1 –j)*j = Σ ((n+1 -j +1)*j {j=1 until n+1} =

= Σ((n+1 –j)*j  +j) = Σ(n+1 -j)*j + Σj {j=1 until n+1} =

=s(n) +(n+1)*(n+2)/2 = n*(n+1)*(n+2)/6 +(n+1)*(n+2)/2 =

= 0.5*(n+1)*(n+2) *(n/3 +1) = (n+1)*(n+2) *(n+3)/6; QED;

Answer 2

For n=1: 1*1=1*2*3/6 (true)

Suppose that for n=k, the statement is true:

k.1+(k-1).2+(k-2).3+...+2(k-1)+1.k=k.(... We assume that this is equation1

We want to prove that for n=k+1, the statement is also true.

(k+1).1+(k).2+(k-1).3+...+2(k)+1.(k+1)... We assume that this equation is equation2 although this equation is not already true.

We also know that (k+1)+(k)+(k-1)+...+3+2+1=(k+1)(k+2)/2, Suppose it is equation 3

equation1+equation3=(k+1)+k*2+(k-1)*3+...

And the right hand side is k(k+1)(k+2)/6+3/6=(k+1)(k+2)(k+3)/6

and we get the second equation.

So because the first equation is true, we can prove that the second equation is true, besides, for n=1, the statement is true, We can conclude that the statement is true for all natural numbers