Key Concepts and Formulas

• Partial Fraction Decomposition: A method to express a rational function as a sum or difference of simpler rational functions. For a denominator with distinct linear factors $(ax+b)(cx+d)$, we can write $\frac{P(x)}{(ax+b)(cx+d)} = \frac{A}{ax+b} + \frac{B}{cx+d}$.
• Telescoping Sums: A series where most of the terms cancel out upon summation, leaving only the first and last few terms. This typically occurs when the general term $a_n$ can be expressed in the form $f(n+k) - f(n)$ or $f(n) - f(n-k)$. The sum of $\sum_{n=1}^{N} (f(n+1) - f(n))$ is $f(N+1) - f(1)$.

Step-by-Step Solution

Step 1: Factorize the Denominator We are given the general term of the sequence as $a_n = \frac{-2}{4n^2 - 16n + 15}$. To apply partial fraction decomposition, we first need to factorize the quadratic denominator $4n^2 - 16n + 15$. We look for two linear factors of the form $(2n-p)(2n-q)$. Expanding this gives $4n^2 - 2(p+q)n + pq$. Comparing the coefficients with $4n^2 - 16n + 15$, we have $2(p+q) = 16 \implies p+q = 8$ and $pq = 15$. The numbers that satisfy these conditions are 3 and 5. Thus, the factorization is $(2n-3)(2n-5)$. So, $a_n = \frac{-2}{(2n-3)(2n-5)}$.

Step 2: Apply Partial Fraction Decomposition We express $a_n$ in the form $\frac{A}{2n-3} + \frac{B}{2n-5}$. $\frac{-2}{(2n-3)(2n-5)} = \frac{A}{2n-3} + \frac{B}{2n-5}$ Multiplying both sides by $(2n-3)(2n-5)$, we get: $-2 = A(2n-5) + B(2n-3)$ To find $A$, let $2n-3 = 0 \implies n = 3/2$. Substituting this value: $-2 = A(2(3/2)-5) + B(0) \implies -2 = A(3-5) \implies -2 = -2A \implies A = 1$ To find $B$, let $2n-5 = 0 \implies n = 5/2$. Substituting this value: $-2 = A(0) + B(2(5/2)-3) \implies -2 = B(5-3) \implies -2 = 2B \implies B = -1$ Therefore, the partial fraction decomposition is: $a_n = \frac{1}{2n-3} - \frac{1}{2n-5}$

Step 3: Identify the Telescoping Pattern We want to express $a_n$ in the form $f(n+k) - f(n)$ or $f(n) - f(n-k)$. Let's rewrite $a_n$ as $a_n = \frac{1}{2n-3} - \frac{1}{2n-5}$. Observe the terms in the denominators: $2n-3$ and $2n-5$. The difference between them is $(2n-3) - (2n-5) = 2$. Let $f(x) = \frac{1}{2x-5}$. Then $f(n) = \frac{1}{2n-5}$. And $f(n+1) = \frac{1}{2(n+1)-5} = \frac{1}{2n+2-5} = \frac{1}{2n-3}$. So, $a_n = f(n+1) - f(n)$. This confirms that the series is a telescoping sum.

Step 4: Calculate the Sum We need to calculate the sum $S = a_1 + a_2 + \dots + a_{25} = \sum_{n=1}^{25} a_n$. Using the telescoping form: $S = \sum_{n=1}^{25} (f(n+1) - f(n))$ Writing out the terms: $S = (f(2) - f(1)) + (f(3) - f(2)) + (f(4) - f(3)) + \dots + (f(26) - f(25))$ The intermediate terms cancel out, leaving: $S = f(26) - f(1)$ Now we calculate $f(1)$ and $f(26)$ using $f(n) = \frac{1}{2n-5}$: For $n=1$: $f(1) = \frac{1}{2(1)-5} = \frac{1}{2-5} = \frac{1}{-3} = -\frac{1}{3}$ For $n=26$: $f(26) = \frac{1}{2(26)-5} = \frac{1}{52-5} = \frac{1}{47}$ Substituting these values back into the sum expression: $S = \frac{1}{47} - \left(-\frac{1}{3}\right) = \frac{1}{47} + \frac{1}{3}$ To add these fractions, we find a common denominator, which is $47 \times 3 = 141$: $S = \frac{3}{141} + \frac{47}{141} = \frac{3+47}{141} = \frac{50}{141}$

Common Mistakes & Tips

• Factorization Errors: Always double-check the factorization of the denominator. A small mistake here will lead to an incorrect partial fraction decomposition.
• Sign Errors in Partial Fractions: Be meticulous when solving for the coefficients $A$ and $B$. Errors in signs are common.
• Identifying the Correct Terms in Telescoping Sums: Carefully write out the first few and last few terms of the sum to ensure the correct initial and final terms are identified for the cancellation. The sum is $f(\text{last index}+k) - f(\text{first index})$. In our case, $k=1$ and the sum is from $n=1$ to $n=25$, so the sum is $f(25+1) - f(1) = f(26) - f(1)$.
• Arithmetic with Fractions: Ensure accuracy when adding or subtracting fractions, especially when dealing with negative signs.

Summary The problem requires summing a series whose general term can be decomposed using partial fractions, leading to a telescoping sum. We first factorized the denominator of $a_n$, then performed partial fraction decomposition to express $a_n$ as the difference of two simpler fractions. Recognizing this as a telescoping sum of the form $f(n+1) - f(n)$, we calculated the sum by evaluating $f(26) - f(1)$, which resulted in $\frac{50}{141}$.

The final answer is $\boxed{\frac{50}{141}}$.