Key Concepts and Formulas

• Generalized Arithmetico-Geometric Progression (AGP): A series where each term is a product of a term from an arithmetic progression and a term from a geometric progression. If the arithmetic part is a polynomial of degree $k$, the series is a generalized AGP.
• Sum of an Infinite Geometric Progression (GP): For a GP with first term $a$ and common ratio $r$ ($|r|<1$), the sum is $S = \frac{a}{1-r}$.
• Method for Summing Generalized AGP: Multiply the series by the common ratio of the geometric part and subtract it from the original series. Repeat this process until the series becomes a standard GP.

Step-by-Step Solution

Step 1: Analyze the Numerator and Identify the Series Type The given series is $S = {7 \over 5} + {9 \over {{5^2}}} + {{13} \over {{5^3}}} + {{19} \over {{5^4}}} + ....$. Let's examine the numerators: $7, 9, 13, 19, \dots$. The first differences are: $9-7=2$, $13-9=4$, $19-13=6$. The second differences are: $4-2=2$, $6-4=2$. Since the second differences are constant, the numerator is a quadratic polynomial in $n$. Let the $n$-th term of the numerator be $N_n = An^2 + Bn + C$. The series is a generalized Arithmetico-Geometric Progression. The common ratio of the geometric part is $r = \frac{1}{5}$.

Step 2: Determine the General Term of the Numerator We use the first three terms to find $A, B, C$: For $n=1$: $A(1)^2 + B(1) + C = 7 \implies A+B+C = 7$ (1) For $n=2$: $A(2)^2 + B(2) + C = 9 \implies 4A+2B+C = 9$ (2) For $n=3$: $A(3)^2 + B(3) + C = 13 \implies 9A+3B+C = 13$ (3)

Subtract (1) from (2): $(4A+2B+C) - (A+B+C) = 9-7 \implies 3A+B = 2$ (4) Subtract (2) from (3): $(9A+3B+C) - (4A+2B+C) = 13-9 \implies 5A+B = 4$ (5)

Subtract (4) from (5): $(5A+B) - (3A+B) = 4-2 \implies 2A = 2 \implies A=1$. Substitute $A=1$ into (4): $3(1)+B=2 \implies B=-1$. Substitute $A=1, B=-1$ into (1): $1+(-1)+C=7 \implies C=7$. Thus, the general term of the numerator is $N_n = n^2 - n + 7$. The series can be written as $S = \sum_{n=1}^{\infty} \frac{n^2 - n + 7}{5^n}$.

Step 3: First Subtraction to Reduce the Degree of the Numerator Write out the series $S$: $S = \frac{7}{5} + \frac{9}{5^2} + \frac{13}{5^3} + \frac{19}{5^4} + \dots \quad (*)$ Multiply $S$ by the common ratio $r = \frac{1}{5}$: $\frac{1}{5}S = \frac{7}{5^2} + \frac{9}{5^3} + \frac{13}{5^4} + \dots \quad (**)$ Subtract (**) from (*): $S - \frac{1}{5}S = \frac{4}{5}S = \frac{7}{5} + \left(\frac{9}{5^2} - \frac{7}{5^2}\right) + \left(\frac{13}{5^3} - \frac{9}{5^3}\right) + \left(\frac{19}{5^4} - \frac{13}{5^4}\right) + \dots$ $\frac{4}{5}S = \frac{7}{5} + \frac{2}{5^2} + \frac{4}{5^3} + \frac{6}{5^4} + \dots$ Let $S' = \frac{2}{5^2} + \frac{4}{5^3} + \frac{6}{5^4} + \dots$. Then, $\frac{4}{5}S = \frac{7}{5} + S'$. The series $S'$ has numerators $2, 4, 6, \dots$, which form an arithmetic progression.

Step 4: Second Subtraction to Convert to a Geometric Progression Now, we find the sum of $S'$: $S' = \frac{2}{5^2} + \frac{4}{5^3} + \frac{6}{5^4} + \dots \quad (***)$ Multiply $S'$ by the common ratio $r = \frac{1}{5}$: $\frac{1}{5}S' = \frac{2}{5^3} + \frac{4}{5^4} + \frac{6}{5^5} + \dots \quad (****)$ Subtract (**) from (*): $S' - \frac{1}{5}S' = \frac{4}{5}S' = \frac{2}{5^2} + \left(\frac{4}{5^3} - \frac{2}{5^3}\right) + \left(\frac{6}{5^4} - \frac{4}{5^4}\right) + \dots$ $\frac{4}{5}S' = \frac{2}{5^2} + \frac{2}{5^3} + \frac{2}{5^4} + \dots$ The right-hand side is an infinite geometric progression with first term $a = \frac{2}{5^2} = \frac{2}{25}$ and common ratio $r = \frac{1}{5}$.

Step 5: Sum the Geometric Progression The sum of the geometric progression is: $G = \frac{a}{1-r} = \frac{2/25}{1 - 1/5} = \frac{2/25}{4/5} = \frac{2}{25} \times \frac{5}{4} = \frac{10}{100} = \frac{1}{10}$ So, we have $\frac{4}{5}S' = \frac{1}{10}$.

Step 6: Back-substitute to Find S From $\frac{4}{5}S' = \frac{1}{10}$, we find $S'$: $S' = \frac{1}{10} \times \frac{5}{4} = \frac{5}{40} = \frac{1}{8}$ Now, substitute $S'$ back into the equation from Step 3: $\frac{4}{5}S = \frac{7}{5} + S'$. $\frac{4}{5}S = \frac{7}{5} + \frac{1}{8}$ Find a common denominator for the right side (which is 40): $\frac{4}{5}S = \frac{7 \times 8}{5 \times 8} + \frac{1 \times 5}{8 \times 5} = \frac{56}{40} + \frac{5}{40} = \frac{61}{40}$ Now, solve for $S$: $S = \frac{61}{40} \times \frac{5}{4} = \frac{61 \times 5}{40 \times 4} = \frac{61}{8 \times 4} = \frac{61}{32}$

Step 7: Calculate the Final Answer The question asks for the value of $160S$. $160S = 160 \times \frac{61}{32}$ We can simplify this calculation: $160 = 5 \times 32$. $160S = (5 \times 32) \times \frac{61}{32} = 5 \times 61$ $160S = 305$

Common Mistakes & Tips

• Algebraic Errors: Be meticulous with arithmetic, especially when subtracting series and summing fractions. A single error can lead to an incorrect final answer.
• Index Alignment: When writing out the series and its shifted version for subtraction, ensure that terms with the same power of the common ratio are aligned correctly.
• Identifying Numerator Pattern: Accurately determining the degree of the polynomial in the numerator is crucial. If it's a linear polynomial, only one subtraction is needed. If it's quadratic, two subtractions are required.

Summary The given series is a generalized Arithmetico-Geometric Progression with a quadratic numerator. The sum was found by repeatedly multiplying the series by the common ratio of the geometric part ($\frac{1}{5}$) and subtracting it from the original series. This process reduces the degree of the numerator polynomial in two steps, transforming the series into a sum of a constant term and a geometric progression. The geometric progression was summed, and then back-substitution was used to find the sum of the original series $S$. Finally, $160S$ was calculated.

The final answer is $\boxed{305}$.