A Series Summation Technique

Mathematics Topic

Finding the sum to n terms of a series 

A Series Summation Technique 
An infinite series, a1 + a2 + a3 +…+an .. whose terms are a function of n with  the highest degree of n being r, will have a partial sum, Sn to n terms which can be expressed in the form: x1nr+1 + x2nr +…+xr+1n, where n,r  E |N,  an, xr E |R.
The values of the coefficients, x1,….,xr+1 can be determined by considering the partial sums from n =1 to n=r+1 which form a system of linear equations. Solving the resulting system will produce the values x1 to xr+1.
After the coefficients are determined, a proof by mathematical induction will validate the resulting partial sum for any natural number, n.
Example 1
Find the partial sum to n terms of: 1*2 + 2*3 + 3*4 +….+n(n+1) 
Since the n-th  term  is a polynomial in n of degree 2:  n(n+1) = n2 + n, Sn has a sum  in the form:
x1n3 + x2n2 + x3n.
For n =1, the LHS is 1*2 = 2  =  x1*13 + x2*12 + x3*1                                   (i)
For n =2 , the LHS is 1*2 + 2*3 = 8  = x1*23 + x2*22 + x3*2                        ( ii)
For n =3, the LHS is 1*2 + 2*3 + 3*4 = 20   = x1*33 + x2*32 + x3*3            (iii) 
The above form a system of linear equations with unknowns x1, x2, x3: 
 x1  +   x2   +   x3  =  2 
8x1 +  4x2   +  2x3  = 8 
27x1 + 9x2   +  3x3  = 20.
The unique solution for this system is:  x1 =1/3, x2 = 1, x3 = 2/3.
Therefore, 1*2 + 2*3 + …+ n(n+1)  = 1/3n3 + n2 + 2/3n. 
A URL for solving linear equations is:  http://wims.unice.fr/wims/wims.cgi?session=D6B8B1F043.2&+lang=en&+module=tool%2Flinear%2Flinsolver.en 
You may also solve the system yourself using variable elimination and substitutions.
Now for a proof by mathematical induction that: 1*2 + 2*3 +….+n(n+1)  =  1/3n3 + n2 + 2/3n.
1*2 = 2   and   1/3*13 + 12 + 2/3*1. = 1/3 + 1 + 2/3 = 2.  Hence true for n=1.
Assume true for some n = k, i.e. 1*2 + 2*3 +… + k(k+1)  =  1/3k3 + k2 + 2/3k
To show k implies k+1, add the next term - (k+1)(k+2) to both sides:
1*2 +... + k(k+1) +(k+1)(k+2) = 1/3k3 +k2 +2/3k +(k+1)(k+2                                                
                                                  =  1/3k3 + k2 +  2/3k + k2  +3k +2
                                                  =  1/3k3 + 2k2 +  11/3k +2
Now,
      k3   = (k+1)3 - 3k2  - 3k -1
and  k2  = (k+1)2  - 2k -1 
Substituting we have,    
       = 1/3[(k+1)3 – 3[(k+1)2 – 2k -1] -3k -1]               
                              +2[(k+1)2 - 2k -1] + 11/3k +2      
        = 1/3(k+1)3 - (k+1)2 + 2k +1 - k  - 1/3  
                           + 2(k+1)2 - 4k - 2 + 11/3k +2  
       =  1/3(k+1)3 + (k+1)2 +2/3k +2/3  
       =  1/3(k+1)3 + (k+1)2  + 2/3(k+1)
Therefore, k implies k+1,and therefore 1*2 + 2*3 +….+n(n+1)  =  1/3n3 + n2 + 2/3n  is true for all n E |N  
 
Example 2
Find the sum to n terms of : 13 + 53 + 93 + 133 +….+ (4n- 3)3 
Since the n-th term is a polynomial in n of degree 3, the sum, Sn can be expressed in the form:
x1n4 + x2n3 + x3n2 + x4n. 
For n = 1, the LHS is 13 =1,   and the RHS =  x1*14 + x2*13 + x3*12 + x4*1
For n = 2, the LHS is 13 + 53 = 126,  and  the RHS is  x1*24 + x2*23 + x3*22 + x4*2
For n = 3, the LHS is 13 + 53 + 93 = 855, and the RHS  is   x1*34 + x2*33 + x3*32 + x4*3
For n = 4, the LHS is 13 + 53 + 93 + 133  = 3052,  and the RHS is
x1*44 + x2*43 + x3*42 + x4*4 
The above form a system of linear equations with unknowns x1, x2, x3, x4:
     x1 + x2 + x3 + x4  =  1                      (i)
 16x1 + 8x2 + 4x3 + 2x4  =  126               (ii)
81x1 + 27x2 + 9x3 + 3x4  =  855              (iii)
256x1 + 64x2 + 16x3 + 4x4 =  3052          (iv) 
The unique solution for this system is: x1 = 16, x2 = -16, x3 = -2, x4= 3.
(I used the Linear Solver at the URL referenced in Example 1)
 Therefore 13 + 53 + 93 + 133 +….+ (4n- 3)3 = 16n4 – 16n3 - 2n2  + 3n.
 Now for a proof by mathematical induction that:
13 + 53 + 93 + 133 +….+ (4n- 3)3 = 16n4 – 16n3 - 2n2  + 3n      for all n E |N.
For n = 1,  the LHS is  13  = 1  and the RHS is  16 – 16 – 2 + 3  = 1.
Therefore true for n = 1.
Assume true for some n = k, i.e.
13 + 53 + 93 + 133 +….+ (4k- 3)3 = 16k4 – 16k3 - 2k2  + 3k
To show k implies k +1, add the next term ((4(k+1) – 3)3) to both sides:
13 + 53 + 93 + 133 +….+(4k- 3)3  + 4(k+1)–3)3= 16k4 –16k3 -2k2 +3k + (4(k+1)–3)3                                                                                                                                      (v)
 Now, 
  (4(k+1) – 3)3  = (4k +1)3 
                      =  64k3 + 48k2 + 12k + 1
The RHS of (v) becomes:
   16k4 – 16k3 - 2k2  + 3k + 64k3 + 48k2 + 12k + 1 
 = 16k4 + 48k3   + 46k2 + 15k +1 
Now,
k4 = (k+1)4 – 4k3 – 6k2 – 4k -1
And   16k4  = 16(k+1)4  - 64k3 – 96k2 - 64k - 16
The RHS of (v) becomes:
     16(k+1)4 - 16k3 – 50k2 – 49k -15
Now, k3 = (k+1)3 – 3k2 – 3k -1
And  -16k3   =  - 16(k+1)3   + 48k2 + 48k + 16
Also k2 = (k+1)2 – 2k -1
The RHS of (v) becomes:
     16(k+1)4 - 16(k+1)3   - 2k2 – k + 1
     =16(k+1)4 - 16(k+1)3   - 2[(k+1)2 – 2k -1] - k + 1
     =16(k+1)4 - 16(k+1)3   - 2(k+1)2   +4k +2 - k + 1
     =16(k+1)4 - 16(k+1)3   - 2(k+1)2   + 3k +3
     =16(k+1)4 - 16(k+1)3   - 2(k+1)2   + 3(k +1) 
Therefore k implies k+1
And therefore 13 + 53 + 93 + 133 +….+ (4n- 3)3 = 16n4 – 16n3 - 2n2  + 3n 
  for all n E |N 
Notes:
Some series will have fewer than the r+1 terms in the sum. However, the technique still 
works as the coefficient(s) of  the  "missing" term(s) will calculate to zero.
 Example:  1 + 3 + 5 +… + 2n-1 = n2  +0n = n2                               
For comments or questions, please feel free to contact me. 
Calvin C. Campbell
Email address: cccamper@netzero.net       
More Examples at: 
http://cccamper.mysite.com/seriesPart2.html