Bernoulli Numbers

Most people learn Gauss's way of adding first n numbers in secondry school. Many are also introduced the idea of adding their squares and cubes. These formulas are (n*(n+1)/2), (n*(n+1)*(2n+1))/ 6 and (n*(n+1)/2)^2 respectively. However the higher formulas are seldom talked about unless you are really curious. What about the sum of 4th power of each number, 5th power and so on. Well its pity that the formula to derive these is not difficult and can easily be demonstrated as I do here.
Let S= 1^p+2^p+3^p+ ... +n^p
Let us expand (v+S)^P= v^P+Pv^(P-1)S+P2v^(P-2)S^2+..   eq(1)
Similarly (v+S-1)^P=v^P+Pv^(P-1)(S-1)+P2v^(P-2)(S-1)^2+.. eq(2)
Subtracting (2) from (1) we get

(v+S)^P-(v+S-1)^P= Pv^(P-1)(S-(S-1))+P2v^(P-2)(S^2-(S-1)^2)+P3v^(P-3)(S^3-(S-1)^3)+...  eq(3)

To simplify the equation let  S^2-(S-1)^2 = 0, S^3-(S-1)^3 =0 and so on and we end up with
(v+S)^P-(v+S-1)^P= Pv^(P-1)(S-(S-1))
 (v+S)^P-(v+S-1)^P= Pv^(P-1)  eq(4)
Eq (4) is important let us substitute different values of v starting from 1..n we get
(1+S)^P-(1+S-1)^P= P*1^(P-1)
(2+S)^P-(2+S-1)^P= P*2^(P-1) 
(3+S)^P-(3+S-1)^P= P*3^(P-1)


(n+S)^P-(n+S-1)^P= P*n^(P-1)

Adding we get
(n+S)^P-S^P=P*(1^(P-1)+2^(P-1)+3^(P-1)+...+n^(P-1))  eq(5)
Let (P-1) =p then (5) becomes
(n+S)^P-(S)^P= P*(1^p+2^p+3^p+...+n^p)
or   ((n+S)^P-(S)^P))/P = 1^p+2^p+3^p+...+n^p = S