prove summation for n(n+1)(2n+1) / 6?

Answer 1

n(n + 1)(2n + 1) / 6

n | 1 | 2 | 3 | 4 | 5

1/6 n (n + 1) (2 n + 1) | 1 | 5 | 14 | 30 | 55

Answer 2

do an update and ask to prove { it is trivial using induction concepts } Σ {k = 1 ,2,3..n} k² = n(n+1)(2n+1) / 6....you should try it...

Answer 3

Do you mean prove the summation formula  for  Σ i^2    from i= 1 to n  ?

By  Induction    :  given  Σ i^2 = n(n+1)(2n+1) / 6     prove this to be true.

a)  true  for   i = n = 1   ?     

Σ i^2  = Σ 1^2 = 1*(1+1)(2(1)+1) / 6  = 1*2*3 / 6 = 6/6 = 1      so  yes , True.

b) assume  formula is true up to   n   that is   , assume

 Σ i^2 = n(n+1)(2n+1) / 6   is  true

c )  show true for   n + 1     [ e.g.  is it true for the next term in the summation, the ( n + 1 )^2    ????  ]

Now  Σ i^2    from  i = 1  to  n+1   equals   Σ i^2 = n(n+1)(2n+1) / 6   for  i=1  to n    PLUS   the next term  which would be   ( n + 1 )^2

so  is   n(n+1)(2n+1) / 6   + (n+1)^2   equal to the formula  replacing  the  upper value in the summation , n ,   with   n+1 ?

n(n+1)(2n+1) / 6   + (n+1)^2  =   ( n+ 1)  [  (n)(2n+1) / 6   + (n+1)  ]

    =   [  (n+1)/6 ] [ n(2n+1 ) + 6(n+1) ]  =  [  (n+1)/6 ] [ 2n^2 + n  + 6n + 6 ]  =

  = [  (n+1)/6 ] [ 2n^2+ 7n +6 ]  =  [  (n+1)/6 ] [ (n+ 2 ) ( 2n + 3 ) ]

 =  [  (n+1)/6 ] [ (  ( n+ 1)  + 1 ) ( 2 ( n+ 1 ) + 1 ) ]

 notice  the  formulas  are the same, with  ( n + 1 ) now  in place  of  n  in the orig.  summation formula.  ..... so by induction the   original summation formula is true.

Don't forget to choose a Best Answer....

Answer 4

I don't see anything to prove, sorry