Series Summation Technique - Part 2

Mathematics Topic

More Examples

 

Example  3.

 Find the partial sum to n terms of: 1 + 7 + 17 +…+ 2n2 -1.. As in Example 1 the n-th term is a polynomial of degree 2. Therefore  the  sum can be expressed in the form: 

 x1n3 + x2n2 + x3n.

For n = 1, the LHS is  1 =1. The RHS is x1*13 + x2*12 + x3*1 

For n = 2, the LHS is  1 + 7 = 8. The RHS is x1*23 + x2*22 + x3*2  

For n = 3, the LHS is  1 + 7 + 17 = 25. The RHS is x1*33 + x2*32 + x3*3 

The resulting system of linear equations is:   x1 +  x2  +   x3   =  1        (i), 

                                                              8x1  +  4x2 +  2x3  =  8      (ii),       

                                                             27x1 + 9x2 + 3x3  = 25       (iii).

I will solve this one using variable eliminations and substitutions.

8* (i) - (ii) : 

     8x1 +  8x2  +   8x3   =  8 

 -  8x1 +  4x2  +   2x3   =   8    

 --------------------------------

               4x2  +  6x3   =    0       (iv)

  27* (i) -  (iii) :

   27x1 + 27x2  +   27x3   =  27

-  27x1 +   9x2  +     3x3   =  25         

------------------------------------

              18x2  +    24x3    = 2     (v) 

 (v) - 4*(iv) :                  

         18x2  +    24x3    = 2     

  -      16x2  +    24x3    = 0  

 --------------------------------

          2x2                     = 2         

        =>  x2   = 1

Substituting   in (iv),  

 4 + 6x3   = 0     or x3 =   -2/3

Substituting in (i), 

   x1 + 1 –2/3   = 1  or x1 = 2/3

Therefore x1 =2/3, x2 = 1 and x3 =-2/3, and  1 + 7 + 17 +…+ 2n2 -1 = 2/3n3 + n2  - 2/3n.

Proof by  mathematical induction:

For n =1, the LHS is 1 and the RHS is  2/3*13 + 12  - 2/3*1 = 1

Therefore true for n =1.

Assume true for n=k,

1 + 7 + 17 +…+ 2k2 -1 = 2/3k3 +k2  - 2/3k

To show k implies k + 1, add 2(k+1)2  - 1 to both sides : 

1 + 7 + 17 +…+ 2k2 -1  + 2(k+1)2= 2/3k3 +k2  - 2/3k +  2(k+1)2  - 1

                                = 2/3[(k+1)3 - 3k2  -3k - 1]  +k2  - 2/3k +  2(k+1)2 - 1

                       = 2/3(k+1)3 - 2k2  -2k - 2/3  +k2  - 2/3k +  2(k+1)2 - 1

                               = 2/3(k+1)3 - k2   -8/3k - 5/3  +  2(k+1)2 

                                = 2/3(k+1)3 - [(k+1)2 -2k -1]  -8/3k  +  2(k+1)2  - 5/3 

                                = 2/3(k+1)3 -(k+1)2 + 2k +1  -8/3k  +  2(k+1)2  - 5/3          

                        = 2/3(k+1)3 +(k+1)2  - 2/3k  - 2/3        

                        = 2/3(k+1)3 +(k+1)2  - 2/3(k +1)

Therefore, k implies k+1, and hence true for all n E |N.    

Example 4

 

What is the sum of  1 + 3 + 6 + ....+ n(n+1)/2 ?

 

This is the sum of sums  problem represented in the Christmas song:  ‘On the first day of Christmas my true love gave to me..'

 

Since the n-th term is a  polynomial  of degree 2  ( n(n+1)/2 = 1/2n2 + 1/2n) ), the sum can be expressed in the form:

 

  x1n3 + x2n2 + x3n.

 

For n = 1,  the LHS is 1 = x1*13 + x2*12 + x3*1

For n = 2,  the LHS is 4  = x1*23 + x2*22  + x3*2

For n = 3, the LHS is 10 = x1*33 + x2*32  + x3*3

 

The  resulting system of  linear of  equations is:

 

       x1  +  x2  +  x3      = 1              (i)

     8x1  + 4x2  + 2x3    = 4              (ii)

   27x1  + 9x2  + 3x3      = 10           (iii)

 

The unique solution for this system is x1 =1/6, x2 = 1/2, and x3 = 1/3

(I used the Linear Solver referenced in Example 1)

 

Therefore 1 + 3 + 6 +...+ n(n+1)/2 = 1/6n3 + 1/2n2 + 1/3n.

 

Proof by mathematical induction:

 

 

For n = 1, we have 1 = 1/6*13 + 1/2*12  + 1/3*1 =1. Therefore, true for n = 1.

 

Assume true for n =  k,

 

1 + 3 + 6 +....+ k(k +1)/2  = 1/6k3 + 1/2k2 + 1/3k.

 

Adding (k +1)(k + 2)/2 to both sides we get,

 

1 + 3 + 6 +...+ k(k +1)/ 2 +(k +1 )(k +2)/2 = 1/6k3 + 1/2k2 + 1/3k  + (k +1)(k +2)/2

                                                                      =  1/6k3 + 1/2k2 + 1/3k  1/2k2 + 3/2k + 1

                                                                      =  1/6k3 + k2  + 11/6k + 1

                                                                      = 1/6[(k+1)3 -3k2 - 3k -1] + k2 + 11/6k + 1

                                                                      =  1/6(k +1)3 - 1/2k2 - 1/2k -1/6 + k2 + 11/6k + 1

                                                                      =  1/6(k+1)3 + 1/2k2 + 4/3k +5/6

                                                                      =  1/6(k+1)3 + 1/2[(k +1)2 -2k -1] + 4/3k + 5/6

                                                                      =  1/6(k+1)3 + 1/2(k+1)2  - k - 1/2 + 4/3k + 5/6

                                                                       = 1/6(k+1)3 + 1/2(k+1)2  + 1/3k  + 1/3

                                                                       = 1/6(k+1)3 + 1/2(k+1)2 + 1/3(k +1).

Therefore k implies k +1,

and hence 1 + 3 + 6 +...+ n(n+1)/2 = 1/6n3 + 1/2n2 + 1/3n is  true for all n  E |N

 

By the way, if you were lucky enough to have been the recipient of the presents in the Christmas song mentioned, over the 12 days you would have received a total of  : 1/6*123 + 1/2*122 + 1/3*12 = 364 presents from your true love!

Again, for questions or comments please feel free to email:

cccamper@netzero.net