MathematicsEasy178×since 2002Q5336The sum ∑n=1213(4n−1)(4n+3)\sum\limits_{n = 1}^{21} {{3 \over {(4n - 1)(4n + 3)}}}n=1∑21(4n−1)(4n+3)3 is equal toA787\frac{7}{87}877B729\frac{7}{29}297C1487\frac{14}{87}8714D2129\frac{21}{29}2921Check answerSkip