M2 MI問題......thx~~~~

1*n+2*(n－1)+3*(n－2)+....+(n－1)*2+n*1=n(n+1)(n+2)/6
求完整的數學歸納法
Sol
當n=1時
左=1*1=1
右=1*2*3/6=1
So n=1時為真
設 n=k時為真即
1*k+2*(k－1)+3*(k－2)+....+(k－1)*2+k*1=k(k+1)(k+2)/6
So
1*(k+1)+2*k+3*(k－1)+....+k*2+(k+1)*1
=[1*k+2*(k－1)+3*(k－2)+....+(k－1)*2+k*1]+[1*1+2*1+3*1+…+k*1]+(k+1)*1
=k(k+1)(k+2)/6+k(k+1)/2+(k+1)*6/6
=[(k+1)/6]*[k(k+2)+3k+6]
=[(k+1)/6]*(k^2+5k+6)
=(k+1)(k+2)(k+3)/6
So
n=k+1時為真

Let P(n) be the require proposition "1n+2(n-1)+3(n-2)+....+(n-1)2+n1=n(n+1)(n+2)/6"
When n = 1,
LHS = 1*1 = 1
RHS = 1*2*3/6 =1
LHS = RHS
P(1) is true.

Assume P(k) is true, where k is any positive integer.
i.e. 1k+2(k-1)+3(k-2)+....+(k-1)2+k1=k(k+1)(k+2)/6

when n = k+1,
LHS = 1(k+1)+2k+3(k-1)+....+k2+(k+1)1
= [1k+2(k-1)+3(k-2)+....+(k-1)2+k1]+[1*1+2*1+3*1+…+k*1]+(k+1)*1
= k(k+1)(k+2)/6+k(k+1)/2+(k+1)
= (k+1)/6*[k(k+2)+3k+6]
= (k+1)/6*(k^2+5k+6)
= (k+1)(k+2)(k+3)/6
RHS = (k+1)(k+1+1)(k+1+2)/6
= (k+1)(k+2)(k+3)/6
LHS=RHS
P(k+1) is also true.
By the principal of Mathematical Induction, P(n) is true for all positive integers n.
1n+2(n-1)+3(n-2)+....+(n-1)2+n1=n(n+1)(n+2)/6 for all positive integers n.