The sum of the infinite series 1 + 5/6 + 12/6² + 22/6³ + 35/6⁴ + 51/6⁵ + 70/6⁶ + ... is equal to:

Q: Wait, why do we subtract twice? Can't we just find the general term $T_n$ and use summation formulas?

You absolutely could use the general term $T_n$ and the method of differences, but that path is often algebraically treacherous and prone to errors. The 'shift and subtract' method is the elegant, standard approach for AGPs. We subtract twice because the second differences of the numerators are constant. Each subtraction effectively 'peels' away one layer of the polynomial growth in the numerator, eventually leaving you with a simple, beautiful geometric series.

Q: What if the series wasn't infinite? Would this method still work?

Yes, it works perfectly for finite series as well! However, you must be extremely careful with the last term. In a finite series, the 'shifted' series will have one extra term at the end that doesn't get subtracted, which you must account for. For infinite series, the terms approach zero, which makes the calculation much cleaner.

Q: I'm getting confused with the alignment during subtraction. Any tips?

Alignment is everything! When you write out the series $S$ and the shifted series $\frac{S}{6}$, physically write them one above the other on your paper, ensuring that terms with the same power of $6$ in the denominator are in the same column. If you don't align them, you will almost certainly miss a term or miscalculate the subtraction. Take your time—precision is the hallmark of a JEE topper.

Question

The sum of the infinite series $1 + \frac{5}{6} + \frac{12}{6^2} + \frac{22}{6^3} + \frac{35}{6^4} + \frac{51}{6^5} + \frac{70}{6^6} + \dots$ is equal to:

Options:

(A) $\frac{425}{216}$

(B) $\frac{429}{216}$

(C) $\frac{288}{125}$

(D) $\frac{280}{125}$

Sigma Percentile · Accepted Answer

### The Beauty of the Infinite Dance

Welcome, fellow traveler of the mathematical landscape. Today, we stand before a series that seems to grow in complexity with every term:

$$S = 1 + \frac{5}{6} + \frac{12}{6^2} + \frac{22}{6^3} + \frac{35}{6^4} + \dots$$

At first glance, it is a chaotic dance of numbers. But look closer. In mathematics, chaos is often just a pattern we haven't decoded yet. Let us embark on a journey to unravel this mystery.

### Phase 1: Decoding the Numerators

Every infinite series has a heartbeat, and for this one, it lies in the numerators: $1, 5, 12, 22, 35, \dots$. Let us act as detectives.

What happens when we look at the gaps between these numbers? The first differences are $5-1=4$, $12-5=7$, $22-12=10$, and $35-22=13$.

We see a sequence: $4, 7, 10, 13, \dots$. This is an Arithmetic Progression (AP) with a common difference of $3$.

If we look at the second differences—the gaps between $4, 7, 10, 13$—we find $3, 3, 3$. The second difference is constant! This tells us we are dealing with a second-order Arithmetico-Geometric Progression (AGP).

The strategy is clear: we must multiply by the common ratio, $r = \frac{1}{6}$, and subtract twice to strip away the complexity.

### Phase 2: The First Transformation

Imagine you are standing on a bridge, watching the terms of the series flow by. We take our original sum $S$ and multiply it by $\frac{1}{6}$.

When we write this out, we shift every term one position to the right to align the denominators:

$$\frac{S}{6} = \frac{1}{6} + \frac{5}{6^2} + \frac{12}{6^3} + \dots$$

Now, we perform the first subtraction: $S - \frac{S}{6}$. On the left, we get $\frac{5S}{6}$.

On the right, the first term $1$ remains, and we subtract the aligned terms:

$$\frac{5S}{6} = 1 + \left(\frac{5}{6} - \frac{1}{6}ight) + \left(\frac{12}{6^2} - \frac{5}{6^2}ight) + \dots$$

This simplifies to:

$$\frac{5S}{6} = 1 + \frac{4}{6} + \frac{7}{6^2} + \frac{10}{6^3} + \dots$$

Let us call this new, slightly tamed series $S_1$.

### Phase 3: The Second Transformation

Look at $S_1$. The numerators are now $1, 4, 7, 10, \dots$. We have successfully reduced the order of the series!

It is now a standard first-order AGP. We repeat our magic trick. We multiply $S_1$ by $\frac{1}{6}$ and shift again:

$$\frac{S_1}{6} = \frac{1}{6} + \frac{4}{6^2} + \frac{7}{6^3} + \dots$$

Subtracting this from $S_1$ gives us:

$$\frac{5S_1}{6} = 1 + \left(\frac{4}{6} - \frac{1}{6}ight) + \left(\frac{7}{6^2} - \frac{4}{6^2}ight) + \dots$$

This simplifies to:

$$\frac{5S_1}{6} = 1 + \frac{3}{6} + \frac{3}{6^2} + \frac{3}{6^3} + \dots$$

Do you see it? The constant numerator of $3$ has emerged like a diamond from the rough.

### Phase 4: The Grand Finale

We are left with a beautiful, infinite Geometric Progression: $\frac{3}{6} + \frac{3}{6^2} + \frac{3}{6^3} + \dots$. The first term $a$ is $\frac{1}{2}$ and the common ratio $r$ is $\frac{1}{6}$.

Using the sum formula $S_{\infty} = \frac{a}{1-r}$, we get:

$$S_{\infty} = \frac{\frac{1}{2}}{1 - \frac{1}{6}} = \frac{\frac{1}{2}}{\frac{5}{6}} = \frac{3}{5}$$

Now, we backtrack. Our equation was $\frac{5S_1}{6} = 1 + \frac{3}{5} = \frac{8}{5}$. Solving for $S_1$, we get:

$$S_1 = \frac{8}{5} 	imes \frac{6}{5} = \frac{48}{25}$$

Finally, recall that $\frac{5S}{6} = S_1$. Thus, $\frac{5S}{6} = \frac{48}{25}$, which leads us to:

$$S = \frac{48}{25} 	imes \frac{6}{5} = \frac{288}{125}$$

We have conquered the series. Remember, in JEE, it is not just about the answer; it is about the elegance of the process. You have just mastered the second-order AGP. The final answer is **$\frac{288}{125}$**.

Animated Solution for Mathematics - Sequence and Series: The sum of the infinite series $1 + \frac{5}{6} + \frac{12}{6 ^{2}} + \frac{22}{6 ^{3}} + \frac{35}{6 ^{4}} + \frac{51}{6 ^{5}} + \frac{70}{6 ^{6}} + \dots$ is equal to:

Select Answer:

Visualized Solution

Analyze the Series Structure

First Differences of Numerators

Second Differences and Strategy

First Shift and Subtract

Result of First Subtraction

Second Shift and Subtract

Result of Second Subtraction

Identify the Infinite GP

Calculate the GP Sum

Solve for $S_{1}$

Substitute back for $S$

Final Calculation

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

The Beauty of the Infinite Dance

Phase 1

Phase 2

Phase 3

Phase 4

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