mathematical Induction

Let P(n) be the proposition
' 1(n)+2(n-1)+3(n-2)+...+2(n-1)+1(n)=n(n+1)(n+2)/6 '

when n= 1
L.H.S = 1(1)
= 1
R.H.S = 1(1+1)(1+2)/6
= 1
becouse, L.H.S=R.H.S
so , P(1) is true

Assume that P(k) is true
i.e. ' 1(k)+2(k-1)+3(k-2)+...+2(k-1)+1(k)=k(k+1)(k+2)/6 '

when n= k +1
L.H.S = 1(k)+2(k-1)+3(k-2)+...+2(k-1)+1(k)+1+2+... k+K+1
=k(k+1)(k+2)/6+1/2(k+2)(k+1)
=(k+1)(k+2)(k+3)/6
= R.H.S
because L.H.S= R.H.S
so, by the principal of mathematical induction , P(n) is true forall natural numbers n.

缺乏解釋

1(k+1)+2(k)+3(k-1)+. ..+2(k)+1(k+1)
=1(k)+2(k-1)+3(k-2)+. ..+2(k-1)+1(n)+1+2+...+k+k+1
=k(k+1)(k+2)/6+1/2(k+2)(k+1)
=1/6(k+1)(k+2)(k+3)