Question

Find the value of:
\[ \frac{5}{2^2 \cdot 3^2} + \frac{7}{3^2 \cdot 4^2} + \frac{9}{4^2 \cdot 5^2} + \frac{11}{5^2 \cdot 6^2} + \frac{13}{6^2 \cdot 7^2} + \frac{15}{7^2 \cdot 8^2} \]

• A. $\dfrac{1}{64}$
• B. $\dfrac{15}{64}$
• C. $\dfrac{15}{16}$
• D. $\dfrac{7}{64}$

Hint:

Whenever you see expressions of the form \( \frac{2n+1}{n^2(n+1)^2} \), try rewriting the numerator using the identity \( (n+1)^2 - n^2 \). This converts the series into a telescoping one.

Answer 1

The Correct Option is B

Step 1: Observe the general pattern of the terms.
Each denominator is of the form \( n^2 (n+1)^2 \).
Each numerator follows the pattern \( 2n + 1 \).
Step 2: Use algebraic identity.
We know that:
\[ (n+1)^2 - n^2 = 2n + 1 \]
Step 3: Rewrite each term using the identity.
\[ \frac{2n+1}{n^2 (n+1)^2} = \frac{(n+1)^2 - n^2}{n^2 (n+1)^2} \]
Step 4: Split each term into partial fractions.
\[ = \frac{1}{n^2} - \frac{1}{(n+1)^2} \]
Step 5: Rewrite the entire series.
\[ \left(\frac{1}{2^2} - \frac{1}{3^2}\right) + \left(\frac{1}{3^2} - \frac{1}{4^2}\right) + \left(\frac{1}{4^2} - \frac{1}{5^2}\right) \]
\[ + \left(\frac{1}{5^2} - \frac{1}{6^2}\right) + \left(\frac{1}{6^2} - \frac{1}{7^2}\right) + \left(\frac{1}{7^2} - \frac{1}{8^2}\right) \]
Step 6: Apply telescoping cancellation.
All intermediate terms cancel out.
Remaining terms are:
\[ \frac{1}{2^2} - \frac{1}{8^2} \]
Step 7: Simplify the remaining expression.
\[ \frac{1}{4} - \frac{1}{64} \]
\[ = \frac{16 - 1}{64} \]
\[ = \frac{15}{64} \]
Step 8: Final conclusion.
Hence, the value of the given expression is $\dfrac{15{64}$}.