Finding an expression for the sum of n tems of the series 1 2 + 2 2 + 3 2 + . . . + n 2 I know that if you have a non-arithmetic or geometric progression, you can find a sum S of a series with the formula S = f ( n + 1 ) − f ( 1 ) where the term u n is u n = f ( n + 1 ) − f ( n ). Then you can prove that with induction.What I don't understand is how I should go about finding the function f(n). For example if I want to calculate the sum to n terms of the series 1 2 + 2 2 + 3 2 + . . . + n 2 then, according to my textbook, my f(n) function should be a polynomial with degree one more than the degree of a term in my sequence - so because the nth term in the sequence is P ( n ) = n 2 then the function f(n) should be f ( n ) = a n 3 + b n 2 + c n + d. But how did they know that it should look like that and how do I gain some intuition into finding that function to help me solve similar problems in the future?

Question

Finding an expression for the sum of n tems of the series       1    2    +      2    2    +      3    2    +  .  .  .  +      n    2  I know that if you have a non-arithmetic or geometric progression, you can find a sum S of a series with the formula   S  =  f  (  n  +  1  )  −  f  (  1  ) where the term       u    n   is       u    n    =  f  (  n  +  1  )  −  f  (  n  ). Then you can prove that with induction.What I don&#039;t understand is how I should go about finding the function f(n). For example if I want to calculate the sum to n terms of the series       1    2    +      2    2    +      3    2    +  .  .  .  +      n    2   then, according to my textbook, my f(n) function should be a polynomial with degree one more than the degree of a term in my sequence - so because the nth term in the sequence is   P  (  n  )  =      n    2   then the function f(n) should be   f  (  n  )  =  a      n    3    +  b      n    2    +  c  n  +  d. But how did they know that it should look like that and how do I gain some intuition into finding that function to help me solve similar problems in the future?

tektonikafrs · Accepted Answer

Step 1This works for sums of pth powers of k because of the fact that   (  n  +  1      )          p      +      1        −      n          p      +      1        ,, when expanded by the binomial theorem, will have no       n          p      +      1       term, and so when summed only uses powers up to the pth power. Also before expanding it, its sum &quot;telescopes&quot; (all terms cancel but two, or all but one if you sum starting at 0.).Step 2Also once you accept the fact you can use the first few values of the sum to determine the constants in front of the powers, as in the a,b,c,d of your example, by solving a linear system.

Finding an expression for the sum of n tems of the series 1^2+2^2+3^2+...+n^2

Answered question

Answer & Explanation

New Questions in Algebra II