Problem 7 Sums Series

Problem 7: Sums & Series

Art of Problem Solving

Problem: Find the value of 12 22 5002 + + ... + 1·3 3·5 999 · 1001 Solution: We can express this in sigma notation: 500 X n=1

n2 (2n + 1)(2n − 1)

We now use a technique called Partial Fraction Decomposition. But before we can use PFD, we need to get rid of the numerator. We can do this by expanding the denominator and then using simple polynomial division. We get n2 1 1/4 = + (2n + 1)(2n − 1) 4 (2n + 1)(2n − 1) The whole point is to make things telescope, and to do that, we use PFD: A B A(2n − 1) + B(2n + 1) 1/4 = + = (2n + 1)(2n − 1) 2n + 1 2n − 1 (2n + 1)(2n − 1) Equating the numerators, we get A(2n − 1) + B(2n + 1) = 2n(A + B) + (B − A) = Since are no n terms on the RHS, A + B = 0. Then B − A = Solving, we get A =

−1 8

and B =

1 4

1 4

1 8

Now we go back to our original sum: 500 X n=1

500

500

n=1

n=1

X1 X −1/8 n2 −1/8 1/8 1/8 = + + = 125 + + (2n + 1)(2n − 1) 4 2n + 1 2n − 1 2n + 1 2n − 1

Looking at our new sum, we see all the terms will cancel out, or telescope, except for the first one and the last one. So we have a final answer of 125 +

1/8 1/8 125250 − = 1 1001 1001

.

Solution was written by Sean Soni and compiled from Art of Problem Solving Forums.