# Evaluate the sum 1/1*3+1/3*5+...+1/(2n-1)(2n+1).

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We notice that we can write the k-th term of the sum as:

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

We'll decompose the term in a sum of elementary fractions:

1/(2k-1)(2k+1) = A/(2k-1) + B/(2k+1)

We'll factorize by (2k-1)(2k+1) both sides:

1 = A(2k+1) + B(2k-1)

We'll remove the brackets:

1 = 2Ak + A + 2Bk - B

1 = 2k(A+B) + (A-B)

Comparing both sides, we'll get:

A+B = 0 => A = -B

A-B = 1

A = 1 + B

1 + 2B = 0

B = -1/2 => A = 1/2

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

Now, we'll calculate the sum:

k = 1=> 1/1*3 = 1/2*1 - 1/2*3

k = 2 => 1/3*5 = 1/2*3 - 1/2*5

......................................

k = n => 1/(2n-1)(2n+1) = 1/2*(2n-1) - 1/2*(2n+1)

We'll add the terms, we'll eliminate the like ones and we'll get:

1/1*3 +...+ 1/(2n-1)(2n+1) = (1/2)[1 - 1/(2n+1)]

1/1*3 +...+ 1/(2n-1)(2n+1) = (1/2)[(2n + 1 - 1)/(2n+1)]

We'll simplify and we'll get:

**1/1*3 +...+ 1/(2n-1)(2n+1) = n/(2n+1)**