Key Concepts and Formulas

• Difference of Squares: $a^2 - b^2 = (a-b)(a+b)$
• Partial Fraction Decomposition: Expressing a rational function as a sum or difference of simpler rational functions. For terms of the form $\frac{1}{x(x+1)}$, the decomposition is $\frac{1}{x} - \frac{1}{x+1}$.
• Telescoping Series: A series where most terms cancel out upon summation, leaving a simple expression.

Step-by-Step Solution

Step 1: Identify the General Term and the Range of the Series The given series is $S = {1 \over {{3^2} - 1}} + {1 \over {{5^2} - 1}} + {1 \over {{7^2} - 1}} + .... + {1 \over {{{(201)}^2} - 1}}$. The denominators are of the form $(odd \ number)^2 - 1$. The odd numbers are $3, 5, 7, \dots, 201$. We can represent these odd numbers as $2k+1$. For the first term, $2k+1=3 \implies k=1$. For the last term, $2k+1=201 \implies 2k=200 \implies k=100$. Thus, the series has $100$ terms, and the general term $T_k$ can be written as: $T_k = {1 \over {{{(2k+1)}^2} - 1}}$ where $k$ ranges from $1$ to $100$.

Why this step? Identifying the general term and the number of terms is fundamental to solving any series problem. It allows us to work with a single formula that represents all terms and to define the limits of summation.

Step 2: Simplify the Denominator using the Difference of Squares Identity The denominator of $T_k$ is $(2k+1)^2 - 1^2$. Using the difference of squares identity, $a^2 - b^2 = (a-b)(a+b)$, with $a = 2k+1$ and $b=1$: $(2k+1)^2 - 1^2 = ((2k+1) - 1)((2k+1) + 1)$ $= (2k)(2k+2)$ Factoring out a 2 from the second term: $= (2k) \cdot 2(k+1)$ $= 4k(k+1)$ So, the general term becomes: $T_k = {1 \over {4k(k+1)}}$

Why this step? Simplifying the denominator into a product of factors is crucial for enabling partial fraction decomposition, which is the key to creating a telescoping series.

Step 3: Decompose the General Term using Partial Fractions We need to decompose the term $\frac{1}{k(k+1)}$. Let: ${1 \over {k(k+1)}} = {A \over k} + {B \over {k+1}}$ Multiplying both sides by $k(k+1)$: $1 = A(k+1) + Bk$ Setting $k=0$, we get $1 = A(1) + B(0) \implies A=1$. Setting $k=-1$, we get $1 = A(0) + B(-1) \implies 1 = -B \implies B=-1$. Thus, ${1 \over {k(k+1)}} = {1 \over k} - {1 \over {k+1}}$ Substituting this back into $T_k$: $T_k = {1 \over 4} \left( {1 \over k} - {1 \over {k+1}} \right)$

Why this step? Expressing each term as a difference of two fractions is the core of creating a telescoping sum. This form ensures that intermediate terms will cancel out during the summation process.

Step 4: Sum the Series (Telescoping Summation) We need to find the sum $S = \sum_{k=1}^{100} T_k$. $S = \sum_{k=1}^{100} {1 \over 4} \left( {1 \over k} - {1 \over {k+1}} \right)$ Factor out the constant $\frac{1}{4}$: $S = {1 \over 4} \sum_{k=1}^{100} \left( {1 \over k} - {1 \over {k+1}} \right)$ Now, let's write out the terms of the sum: $S = {1 \over 4} \left[ \left( {1 \over 1} - {1 \over 2} \right) + \left( {1 \over 2} - {1 \over 3} \right) + \left( {1 \over 3} - {1 \over 4} \right) + \dots + \left( {1 \over {99}} - {1 \over {100}} \right) + \left( {1 \over {100}} - {1 \over {101}} \right) \right]$ In this sum, the second part of each term cancels with the first part of the next term. For example, $-{1 \over 2}$ cancels with $+{1 \over 2}$, $-{1 \over 3}$ cancels with $+{1 \over 3}$, and so on, until $-{1 \over 100}$ cancels with $+{1 \over 100}$. The only remaining terms are the first part of the first term and the second part of the last term: $S = {1 \over 4} \left[ {1 \over 1} - {1 \over {101}} \right]$

Why this step? This is where the telescoping effect is realized. By writing out the terms, we can clearly see the cancellation of intermediate terms, simplifying the summation to just the initial and final components.

Step 5: Calculate the Final Sum Now, we simplify the expression obtained in Step 4: $S = {1 \over 4} \left[ {{101 - 1} \over {101}} \right]$ $S = {1 \over 4} \left[ {{100} \over {101}} \right]$ $S = {{100} \over {4 \times 101}}$ $S = {{25} \over {101}}$

Common Mistakes & Tips

• Forgetting the Constant Factor: When decomposing $\frac{1}{4k(k+1)}$, remember to factor out the $\frac{1}{4}$ before or after the partial fraction decomposition of $\frac{1}{k(k+1)}$.
• Incorrectly Identifying the Number of Terms: Carefully determine the range of the index $k$ to ensure the correct number of terms are included in the summation.
• Algebraic Errors: Double-check all algebraic manipulations, especially during the difference of squares expansion and partial fraction calculation.

Summary

The problem requires summing a series whose terms have denominators that can be simplified using the difference of squares identity. The general term of the series was found to be $T_k = \frac{1}{4k(k+1)}$. By applying partial fraction decomposition, $T_k$ was rewritten as $\frac{1}{4}\left(\frac{1}{k} - \frac{1}{k+1}\right)$. Summing this expression from $k=1$ to $k=100$ revealed a telescoping series, where intermediate terms cancel out. This left the first part of the first term and the last part of the last term, leading to the simplified sum of $\frac{25}{101}$.

The final answer is $\boxed{{25} \over {101}}$, which corresponds to option (B).