Key Concepts and Formulas

• Telescoping Series (Method of Differences): A series where the sum can be simplified by the cancellation of intermediate terms. If the general term $T_r$ can be expressed as $T_r = f(r) - f(r+1)$ (or a similar form), then the sum $\sum_{r=1}^{N} T_r = f(1) - f(N+1)$.
• Algebraic Factorization: Techniques like the difference of squares ($a^2 - b^2 = (a-b)(a+b)$) and completing the square are crucial for manipulating denominators.
• Partial Fraction Decomposition: Used to break down complex fractions into simpler ones, often in the form $\frac{A}{g(r)} + \frac{B}{h(r)}$.

Step-by-Step Solution

Step 1: Determine the General Term ($T_r$) of the Series

The given series is ${1 \over 5} + {2 \over {65}} + {3 \over {325}} + {4 \over {1025}} + {5 \over {2501}} + \,\,....$ We need to find a formula for the $r$-th term, $T_r$.

• Numerator: The numerators are $1, 2, 3, 4, 5, \dots$. This is an arithmetic progression, so the numerator of the $r$-th term is $r$.
• Denominator: The denominators are $5, 65, 325, 1025, 2501, \dots$. Let's examine these values. We can observe that the denominators are related to squares. $5 = 2^2 + 1$ $65 = 8^2 + 1$ $325 = 18^2 + 1$ $1025 = 32^2 + 1$ $2501 = 50^2 + 1$ The sequence of bases is $2, 8, 18, 32, 50, \dots$. Let's find the pattern for this sequence, say $X_r$. $X_1 = 2 = 2 \times 1^2$ $X_2 = 8 = 2 \times 2^2$ $X_3 = 18 = 2 \times 3^2$ $X_4 = 32 = 2 \times 4^2$ $X_5 = 50 = 2 \times 5^2$ Thus, $X_r = 2r^2$. The denominator of the $r$-th term is $(X_r)^2 + 1 = (2r^2)^2 + 1$.

Therefore, the general term of the series is: $T_r = {r \over {(2r^2)^2 + 1}} = {r \over {4r^4 + 1}}$

Step 2: Express $T_r$ as a Difference of Two Terms (Telescoping Form)

We need to rewrite $T_r$ in the form $k \cdot (f(r) - f(r+1))$. First, let's factor the denominator $4r^4 + 1$. This is a standard factorization technique. $4r^4 + 1 = (2r^2)^2 + 1^2$ We add and subtract $2(2r^2)(1) = 4r^2$ to complete the square: $4r^4 + 1 = (2r^2)^2 + 2(2r^2)(1) + 1^2 - 4r^2$ $4r^4 + 1 = (2r^2 + 1)^2 - (2r)^2$ Using the difference of squares formula, $A^2 - B^2 = (A-B)(A+B)$: $4r^4 + 1 = (2r^2 + 1 - 2r)(2r^2 + 1 + 2r)$ $4r^4 + 1 = (2r^2 - 2r + 1)(2r^2 + 2r + 1)$

Now, substitute this back into $T_r$: $T_r = {r \over {(2r^2 - 2r + 1)(2r^2 + 2r + 1)}}$ Observe the difference between the two factors in the denominator: $(2r^2 + 2r + 1) - (2r^2 - 2r + 1) = 4r$ Our numerator is $r$. To make the numerator equal to the difference of the denominator factors, we multiply the numerator and denominator by 4: $T_r = {1 \over 4} \cdot {{4r} \over {(2r^2 - 2r + 1)(2r^2 + 2r + 1)}}$ $T_r = {1 \over 4} \left( {{ (2r^2 + 2r + 1) - (2r^2 - 2r + 1) } \over {(2r^2 - 2r + 1)(2r^2 + 2r + 1)}} \right)$ Now, we can split this into two fractions: $T_r = {1 \over 4} \left( {{1 \over {2r^2 - 2r + 1}} - {{1 \over {2r^2 + 2r + 1}}}} \right)$ This is in the form $k \cdot (f(r) - f(r+1))$. Let $f(r) = {1 \over {2r^2 - 2r + 1}}$. To verify $f(r+1)$: $f(r+1) = {1 \over {2(r+1)^2 - 2(r+1) + 1}}$ $f(r+1) = {1 \over {2(r^2 + 2r + 1) - 2r - 2 + 1}}$ $f(r+1) = {1 \over {2r^2 + 4r + 2 - 2r - 2 + 1}}$ $f(r+1) = {1 \over {2r^2 + 2r + 1}}$ This confirms that $T_r = {1 \over 4} (f(r) - f(r+1))$.

Step 3: Calculate the Sum of the First Ten Terms ($S_{10}$)

We need to find the sum of the first ten terms, $S_{10} = \sum_{r=1}^{10} T_r$. $S_{10} = \sum_{r=1}^{10} {1 \over 4} (f(r) - f(r+1))$ $S_{10} = {1 \over 4} \sum_{r=1}^{10} (f(r) - f(r+1))$ This is a telescoping series. The sum expands as: $S_{10} = {1 \over 4} [(f(1) - f(2)) + (f(2) - f(3)) + \dots + (f(10) - f(11))]$ The intermediate terms cancel out, leaving: $S_{10} = {1 \over 4} [f(1) - f(11)]$ Now, we calculate $f(1)$ and $f(11)$:

• $f(r) = {1 \over {2r^2 - 2r + 1}}$
• $f(1) = {1 \over {2(1)^2 - 2(1) + 1}} = {1 \over {2 - 2 + 1}} = {1 \over 1} = 1$
• $f(11) = {1 \over {2(11)^2 - 2(11) + 1}} = {1 \over {2(121) - 22 + 1}} = {1 \over {242 - 22 + 1}} = {1 \over {221}}$

Substitute these values back into the sum: $S_{10} = {1 \over 4} \left( 1 - {1 \over {221}} \right)$ $S_{10} = {1 \over 4} \left( {{221 - 1} \over {221}} \right)$ $S_{10} = {1 \over 4} \left( {220 \over {221}} \right)$ $S_{10} = {55 \over {221}}$

Step 4: Determine $m+n$

We are given that the sum is $\frac{m}{n}$, where $m$ and $n$ are co-prime. We found $S_{10} = {55 \over {221}}$. So, $m = 55$ and $n = 221$. To check if they are co-prime, we find their prime factors: $55 = 5 \times 11$ $221 = 13 \times 17$ Since there are no common prime factors, $m$ and $n$ are co-prime.

We need to find $m+n$: $m+n = 55 + 221 = 276$

Common Mistakes & Tips

• Denominator Pattern Recognition: The factorization of $4r^4+1$ is a key step that may require practice. Recognizing it as a form of Sophie Germain's Identity is helpful.
• Algebraic Manipulation: Errors in expanding squares or applying the difference of squares formula can lead to incorrect $f(r)$ and $f(r+1)$ terms.
• Co-primality Check: Ensure that the final fraction $\frac{m}{n}$ is indeed in its simplest form by checking for common factors between $m$ and $n$.

Summary

The problem involves finding the sum of a series by first determining its general term, which was found to be $T_r = {r \over {4r^4 + 1}}$. This general term was then cleverly manipulated using algebraic factorization to express it in a telescoping form: $T_r = {1 \over 4} ({1 \over {2r^2 - 2r + 1}} - {1 \over {2r^2 + 2r + 1}})$. Applying the method of differences for the sum of the first ten terms resulted in $S_{10} = {55 \over {221}}$. With $m=55$ and $n=221$ being co-prime, their sum $m+n$ is calculated.

The final answer is $\boxed{276}$.