Method of Undetermined Coefficients

 

What?

 

           The method of undetermined coefficients is an old method for finding the sum of a finite series.

 

 

How?

 

           The best way to show how this method works is to use an example. Suppose we like to find the finite sum of the series:

           1×2 + 2×3 + 3×4 + …. + n(n+1).

 

Let       f(n) = 1×2 + 2×3 + 3×4 + …. + n(n+1) º A0 + A1n + A2n2 + A3n3 + ….        (1)

Then          f(n+1) = 1×2 + 2×3 + 3×4 + …. + n(n+1) + (n+1)(n+2)

                                     º A0 + A1(n + 1) + A2(n + 1)2 + A3(n + 1)3 + ….                        (2)

 

(2) – (1),            (n + 1)(n + 2) º A1 + A2 (2n + 1) + A3(3n2 + 3n + 1) + ….              (3)

 

           Equation (3) is an identity and the highest power of n on the L.H.S. is 2, it follows that

A4, A5, ….   are all equal to zero.

 

Equating coefficients on the two sides of the identity of (3):

           n2 :            1 = 3A3,                                                      \ A3 = 1/3

           n1 :            3 = 2A2 + 3A3 = 2A2 + 1                        \ A2 = 1

           n0 :            2 = A1 + A2 + A3 = A1 + 1 + (1/3)       \ A1 = 2/3

 

           \ 1×2 + 2×3 + 3×4 + …. + n(n+1) º A0 + (2/3)n + n2 + (1/3)n3       ….                           (4)

 

Put n = 1 in (4),               1×2 = A0 + (2/3) + 1 + (1/3)          \ A0 = 0

 

           \ 1×2 + 2×3 + 3×4 + …. + n(n+1)       º  (2/3)n + n2 + (1/3)n3

                                                                                 º      n(n+1)(n+2)/3

 

 

More…

 

Can you use the Method of Undetermined Coefficients to check:

 

(1)      12 + 22 + 32 + …. + n2         º      n(n+1)(2n+1)/6

(2)      13 + 33 + 53 + …. + (2n-1)3 º       n2(2n2 – 1)          (Hint: should work out A4)