mathematical induction

Let P(n) be 1．n + 2．(n - 1) + 3．(n - 2) + ... + (n - 2)．3 + (n - 1)．2 + n．1 = (1/6)(n)( n + 1)( n + 2)
When n = 1,

L.H.S.
=1*1
=1

R.H.S.
= (1/6)(1)( 1 + 1)( 1 + 2)
= 1

As L.H.S. = R.H.S.
So P(1) is true.

Assume P(k) is true, i.e. 1．k + 2．(k - 1) + 3．(k - 2) + ... + (k - 2)．3 + (k - 1)．2 + k．1 = (1/6)(k)( k + 1)( k + 2)
When n = k+1,

L.H.S.
=1．(k+1) + 2．(k ) + 3．(k - 1) + ... + (k - 1)．3 + (k )．2 + (k+1)．1
= 1．(k+1) + 2．(k-1+1) + 3．(k - 2 + 1) + ... + (k - 2 + 1)．3 + (k-1+1)．2 + (k+1)．1
= [1．k + 2．(k - 1) + 3．(k - 2) + ... + (k - 2)．3 + (k - 1)．2 + k．1 ]+[1+2+3+...(k+1)]
= (1/6)(k)( k + 1)( k + 2) + (k+1)(k+2)/2
{by the assumption and the formula of the sum of 1+2+3+...+n}
=1/6(k+1)(k+2)(k+3)
=1/6(k+1)(k+1+1)(k+1+2)
= R.H.S.

So P(k+1) is true.

By Mathematical Induction, 1．n + 2．(n - 1) + 3．(n - 2) + ... + (n - 2)．3 + (n - 1)．2 + n．1 = (1/6)(n)( n + 1)( n + 2) is true for all positive integer of n.

1．1 = (1/6)(1)(1+1)(1+2)

assume
1．k + 2．(k - 1) + ... + (k - 1)．2 + k．1 = (1/6)(k)( k + 1)( k + 2)

1．(k+1) + 2．(k) + ... + (k)．2 + (k+1)．1
= [ 1．k + 2．(k - 1) + ... + k．1 ] + [ 1 + 2 + ... + k ] + (k+1)．1
= (1/6)(k)( k + 1)( k + 2) + k(k+1)/2 + (k+1)
= (1/6)(k+1)(k+2)(k+3) by simplying

just give you important steps and you finish other statement, OK ?