Key Concepts and Formulas

• Infinite Geometric Series: The sum of an infinite geometric series $a + ar + ar^2 + \dots$ is given by $S_{\infty} = \frac{a}{1-r}$, provided that the absolute value of the common ratio $|r| < 1$.
• Arithmetic-Geometric Series: A series where terms are a product of a term in an arithmetic progression and a term in a geometric progression. These can often be solved by multiplying the series by the common ratio of the geometric part and subtracting.

Step-by-Step Solution

1. Representing and Decomposing the Series: We are given the equation: $7 = 5+\frac{1}{7}(5+\alpha)+\frac{1}{7^2}(5+2 \alpha)+\frac{1}{7^3}(5+3 \alpha)+\ldots \ldots \ldots \ldots \infty$ Let the given series be $S$. We can decompose $S$ into two separate infinite series by separating the constant term '5' and the term involving $\alpha$: $S = \left(5 + \frac{5}{7} + \frac{5}{7^2} + \frac{5}{7^3} + \dots\right) + \left(\frac{\alpha}{7} + \frac{2\alpha}{7^2} + \frac{3\alpha}{7^3} + \dots\right)$ We are given that $S=7$.

2. Evaluating the First Series (Geometric Series): The first part is a geometric series: $5 + \frac{5}{7} + \frac{5}{7^2} + \frac{5}{7^3} + \dots$. The first term is $a_1 = 5$. The common ratio is $r_1 = \frac{1}{7}$. Since $|r_1| = \frac{1}{7} < 1$, the sum of this infinite geometric series converges. Using the formula $S_{\infty} = \frac{a}{1-r}$: $S_1 = \frac{5}{1 - \frac{1}{7}} = \frac{5}{\frac{6}{7}} = 5 \times \frac{7}{6} = \frac{35}{6}$

3. Evaluating the Second Series (Arithmetic-Geometric Series): The second part is the series: $S_2 = \frac{\alpha}{7} + \frac{2\alpha}{7^2} + \frac{3\alpha}{7^3} + \dots$. This is an arithmetic-geometric series. To find its sum, let's multiply $S_2$ by the common ratio of the geometric part, which is $\frac{1}{7}$: $\frac{1}{7}S_2 = \frac{\alpha}{7^2} + \frac{2\alpha}{7^3} + \frac{3\alpha}{7^4} + \dots$ Now, subtract this equation from the original $S_2$: $S_2 - \frac{1}{7}S_2 = \left(\frac{\alpha}{7} + \frac{2\alpha}{7^2} + \frac{3\alpha}{7^3} + \dots\right) - \left(\frac{\alpha}{7^2} + \frac{2\alpha}{7^3} + \frac{3\alpha}{7^4} + \dots\right)$ Combine like terms: $\frac{6}{7}S_2 = \frac{\alpha}{7} + \left(\frac{2\alpha}{7^2} - \frac{\alpha}{7^2}\right) + \left(\frac{3\alpha}{7^3} - \frac{2\alpha}{7^3}\right) + \dots$ $\frac{6}{7}S_2 = \frac{\alpha}{7} + \frac{\alpha}{7^2} + \frac{\alpha}{7^3} + \dots$ The right-hand side is now an infinite geometric series with first term $a_2 = \frac{\alpha}{7}$ and common ratio $r_2 = \frac{1}{7}$. Since $|r_2| = \frac{1}{7} < 1$, its sum converges: $\frac{6}{7}S_2 = \frac{\frac{\alpha}{7}}{1 - \frac{1}{7}} = \frac{\frac{\alpha}{7}}{\frac{6}{7}} = \frac{\alpha}{6}$ Now, solve for $S_2$: $S_2 = \frac{7}{6} \times \frac{\alpha}{6} = \frac{7\alpha}{36}$

4. Combining the Series Sums and Solving for $\alpha$: We know that the total sum $S = S_1 + S_2 = 7$. Substitute the calculated values of $S_1$ and $S_2$: $\frac{35}{6} + \frac{7\alpha}{36} = 7$ To eliminate the fractions, multiply the entire equation by the least common multiple of the denominators (36): $36 \left(\frac{35}{6}\right) + 36 \left(\frac{7\alpha}{36}\right) = 36(7)$ $6(35) + 7\alpha = 252$ $210 + 7\alpha = 252$ Subtract 210 from both sides: $7\alpha = 252 - 210$ $7\alpha = 42$ Divide by 7: $\alpha = \frac{42}{7} = 6$

Common Mistakes & Tips

• Incorrectly identifying the series: Ensure you correctly separate the constant terms from the terms involving $\alpha$.
• Algebraic errors in subtraction: When subtracting the shifted series, be meticulous with the signs and corresponding terms.
• Arithmetic-Geometric Series Formula: While there's a direct formula for AGP, the method of subtracting the shifted series is fundamental and less prone to memorization errors. This method effectively converts the AGP into a standard GP.

Summary

The problem requires summing an infinite series. By decomposing the series into a standard geometric series and an arithmetic-geometric series, we can find the sum of each part. The geometric series converges to $\frac{35}{6}$. The arithmetic-geometric series is manipulated by subtracting a shifted version of itself to reveal a standard geometric series, leading to a sum of $\frac{7\alpha}{36}$. Equating the total sum to the given value of 7 and solving the resulting linear equation yields the value of $\alpha$.

The final answer is $\boxed{6}$.