If S = 7/5 + 9/5² + 13/5³ + 19/5⁴ + ..., then 160 S is equal to .

Q: Why can't I just use the standard AGP formula?

The standard AGP formula applies when the numerators are in a simple arithmetic progression (e.g., 1, 2, 3...). Here, the numerators are 7, 9, 13, 19... their differences are 2, 4, 6... which is an AP. This is a higher-order series, so we must use the shifting method twice or recognize the binomial expansion form.

Q: How do I know if the series converges?

For an infinite geometric or arithmetico-geometric series to converge, the common ratio $r$ must satisfy $|r| < 1$. Here, our common ratio is $1/5$. Since $|1/5| < 1$, the series converges beautifully to a finite value.

Q: Why did we multiply by $1/5$ specifically?

We multiply by the common ratio of the geometric part of the series. In our series, the denominators are $5, 5^2, 5^3...$. The common ratio is $1/5$. Multiplying by this ratio aligns the terms with the same powers of 5, allowing us to subtract them and simplify the expression.

Question

If $S = \frac{7}{5} + \frac{9}{5^2} + \frac{13}{5^3} + \frac{19}{5^4} + \dots$, then $160 S$ is equal to .

Sigma Percentile · Accepted Answer

### The Infinite Dance of Numbers

Welcome, fellow traveler of the mathematical landscape. Today, we are going to unravel a puzzle that might seem daunting at first glance: an infinite series that refuses to behave like a simple geometric progression.

We are looking at the series:

$$S = \frac{7}{5} + \frac{9}{5^2} + \frac{13}{5^3} + \frac{19}{5^4} + \dots$$

At first, it looks like a chaotic jumble of fractions, but there is a hidden rhythm here waiting to be discovered.

### Phase 1: The Pattern Hunt

Before we dive into the algebra, let us observe the numerators: $7, 9, 13, 19, \dots$. If you look closely, you will see they are not in a simple arithmetic progression.

However, look at the differences between consecutive terms: $9-7=2$, $13-9=4$, $19-13=6$. The differences are $2, 4, 6, \dots$, which is a perfect arithmetic progression.

This is the DNA of our series. It tells us that we are dealing with a second-order arithmetico-geometric series, which we can solve using the **Shifting Method**.

### Phase 2: The Shifting Method

We multiply the entire series $S$ by the common ratio of the denominators, which is $\frac{1}{5}$. When we do this, every term shifts one position to the right:

$$\frac{1}{5} S = \frac{7}{5^2} + \frac{9}{5^3} + \frac{13}{5^4} + \dots$$

Now, place this shifted series directly under the original one and subtract the shifted series from the original. The terms with the same denominators align, yielding:

$$S - \frac{1}{5} S = \frac{7}{5} + \left( \frac{9}{5^2} - \frac{7}{5^2} ight) + \left( \frac{13}{5^3} - \frac{9}{5^3} ight) + \dots$$

This simplifies to:

$$\frac{4}{5} S = \frac{7}{5} + \frac{2}{5^2} + \frac{4}{5^3} + \frac{6}{5^4} + \dots$$

### Phase 3: The Binomial Connection

Now, look at the right-hand side. If we factor out $\frac{2}{5^2}$ from the terms starting from the second one, we get:

$$\frac{4}{5} S = \frac{7}{5} + \frac{2}{5^2} \left( 1 + 2\left( \frac{1}{5} ight) + 3\left( \frac{1}{5} ight)^2 + \dots ight)$$

The bracketed expression is the classic binomial expansion for $(1-x)^{-2} = 1 + 2x + 3x^2 + 4x^3 + \dots$, where $x = \frac{1}{5}$. Since $|x| < 1$, this expansion is valid.

Substituting $x = \frac{1}{5}$, the bracket becomes:

$$\left( 1 - \frac{1}{5} ight)^{-2} = \left( \frac{4}{5} ight)^{-2} = \left( \frac{5}{4} ight)^2 = \frac{25}{16}$$

### Phase 4: The Final Victory

Substituting this back into our equation, we have:

$$\frac{4}{5} S = \frac{7}{5} + \frac{2}{25} 	imes \frac{25}{16}$$

The $25$s cancel out, leaving us with:

$$\frac{4}{5} S = \frac{7}{5} + \frac{1}{8} = \frac{56+5}{40} = \frac{61}{40}$$

Solving for $S$, we find:

$$S = \frac{61}{40} 	imes \frac{5}{4} = \frac{61}{32}$$

The problem asks for $160 S$. Therefore:

$$160 	imes \frac{61}{32} = 5 	imes 61 = \mathbf{305}$$

Animated Solution for Mathematics - Sequence and Series: If $S = \frac{7}{5} + \frac{9}{5 ^{2}} + \frac{13}{5 ^{3}} + \frac{19}{5 ^{4}} + \dots$ , then $160 S$ is equal to .

Enter Numerical Value:

Visualized Solution

Identify the Given Series $S$

Apply the Shifting Method

Subtract the Two Equations

Simplify the Subtraction Result

Factor Out Common Terms

Identify the Binomial Expansion

Substitute and Solve the Bracket

Update the Main Equation

Calculate the Sum of Fractions

Solve for $S$

Final Calculation: $160 S$

The Sigma Insight: Arithmetic-Geometric Progression (A.G.P.)

The Infinite Dance of Numbers

Phase 1

Phase 2

Phase 3

Phase 4

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