Successor of n=n+1

Predecessor of n=n-1

Sum=[n+1]+[n-1]

Plus 1 plus minus 1=

[1]+[-1]=0

[n][n]

=2n

Hope this helps

n! = n * (n-1) !

n!/n = n*(n-1)!/n //divid by n both sides

n!/n = (n-1)!

let n = 1

1!/1 = (1-1)!

1 = 0!

Hence proved that 0! = 1.

nCr + nCr-1 = n!/[r!(n-r)!] + n!/[(r-1)!(n-r+1)!]

= n!/[(r-1)!(n-r)!]*{1/r + 1/n-r+1}

= n!/[(r-1)!(n-r)!]*{[(n-r+1) + r]/[r*(n-r+1)]}

= n!/[(r-1)!(n-r)!]*{(n+1)/r*(n-r+1)]}

= (n+1)!/[r!(n+1-r)!]

= n+1Cr

Assuming you mean the first n counting numbers then:

let S{n} be the sum; then:

S{n} = 1 + 2 + ... + (n-1) + n

As addition is commutative, the sum can be reversed to give:

S{n} = n + (n-1) + ... + 2 + 1

Now add the two versions together (term by term), giving:

S{n} + S{n} = (1 + n) + (2 + (n-1)) + ... + ((n-1) + 2) + (n + 1)

→ 2S{n} = (n+1) + (n+1) + ... + (n+1) + (n+1)

As there were originally n terms, this is (n+1) added n times, giving:

2S{n} = n(n+1)

→ S{n} = ½n(n+1)

The sum of the first n counting numbers is ½n(n+1).

A recursive formula for the factorial is n! = n(n - 1)!. Rearranging gives (n - 1)! = n!/n, Substituting 'n - 1' as 0 -- i.e. n = 1 -- then 0! = 1!/1, which is 1/1 = 1.