The sum of series represented as 1/(1*5) + 1/(9*13) +....+ 1/(221*225) is

Solution:

• Recognize the general term: the nth term is 1/((4n-3)(4n+1)). The last term 221*225 corresponds to 4n-3 = 221, so n = 56 (terms n = 1 to 56).

• Use partial fractions: 1/((4n-3)(4n+1)) = 1/4 * (1/(4n-3) - 1/(4n+1)).

• Sum from n = 1 to 56. The series telescopes: 1/4[(1/1 - 1/5) + (1/5 - 1/9) + ... + (1/221 - 1/225)] = 1/4(1 - 1/225).

• Compute the result: 1/4 * (224/225) = 56/225.

Answer: 56/225