Telescoping Series Sum

Question

The value of the following sum of terms $\frac{5}{2^2 \cdot 3^2} + \frac{7}{3^2 \cdot 4^2} + \frac{9}{4^2 \cdot 5^2} + \frac{11}{5^2 \cdot 6^2} + \frac{13}{6^2 \cdot 7^2} + \frac{15}{7^2 \cdot 8^2}$ is

Accepted Answer

B. The $n$-th term is $\frac{2n+1}{n^2(n+1)^2} = \frac{1}{n^2} - \frac{1}{(n+1)^2}$. Sum from $n=2$ to $7$ telescopes to $\frac{1}{2^2} - \frac{1}{8^2} = \frac{1}{4} - \frac{1}{64} = \frac{15}{64}$.

Answer

\frac{1}{64}

Answer

\frac{15}{64}

Answer

\frac{15}{16}

Answer

\frac{7}{64}

Telescoping Series Sum

Explanation

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