Let 75...57 denote the (r + 2) digit number where the first and the last digits are 7 and the remain

Q: Wait, why do we multiply the sum $S$ by $10$? Is there a rule for this?

This is a classic technique for series that look like geometric progressions but aren't quite. By multiplying by $10$, we shift the decimal representation of each number, which aligns the digits of $T_r$ with $T_{r+1}$. This alignment is what allows the 'telescoping' effect to happen, where the middle terms cancel out, leaving only the boundaries.

Q: Why does $10T_r - T_{r+1}$ always equal $13$? It seems too simple.

It is simple, and that is the beauty of it! Let's look at the digits. $10T_r$ is the number $75\dots570$ (with $r$ fives). $T_{r+1}$ is $75\dots557$ (with $r+1$ fives). When you subtract $75\dots557$ from $75\dots570$, the leading $75\dots5$ parts cancel out, leaving you with $570 - 557 = 13$. The structure is designed to collapse perfectly.

Q: How do I know when to stop the telescoping? Do I just ignore the $T_{99}$ term?

You don't ignore it! The problem asks for an expression involving $T_{99}$. Our telescoping sum gives us $10T_{98}$. We then use the same logic we used for the other terms: $10T_{98} - T_{99} = 13$. This allows us to replace $10T_{98}$ with $T_{99} + 13$, which perfectly bridges the gap to the required form.

Question

Let $75...57$ denote the $(r + 2)$ digit number where the first and the last digits are 7 and the remaining $r$ digits are 5. Consider the sum $S = 77 + 757 + 7557 + \dots + \underbrace{75...57}_{98}$. If $S = \frac{\underbrace{75...57}_{99} + m}{n}$, where $m$ and $n$ are natural numbers less than 3000, then the value of $m + n$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When you first look at the sum $S = 77 + 757 + 7557 + \dots + \underbrace{75\dots57}_{98}$, it is natural to feel overwhelmed. You see a sequence of numbers growing in length, each with an increasing number of fives.

In mathematics, whenever you see a pattern that grows in a predictable way, there is almost always a hidden symmetry waiting to be exploited. Let us define our general term $T_r$ as the number with $r$ fives between two $7$s.

So, $T_0 = 77$, $T_1 = 757$, and so on. Our goal is to find the sum:

$$S = \sum_{r=0}^{98} T_r$$

### The $10S$ Strategy

The secret to solving this lies in the 'shift and subtract' method. Notice that each term $T_r$ is roughly ten times the previous one.

If we multiply the entire sum $S$ by $10$, we get:

$$10S = 10T_0 + 10T_1 + \dots + 10T_{98}$$

When we write this out, $10T_0$ becomes $770$, $10T_1$ becomes $7570$, and so on. Now, look at what happens when we calculate $9S = 10S - S$. We are subtracting the original sum from our shifted sum.

### The Telescoping Magic

Let us examine the difference $10T_r - T_{r+1}$. For $r=0$, we have $10T_0 - T_1 = 770 - 757 = 13$. For $r=1$, we have $10T_1 - T_2 = 7570 - 7557 = 13$.

Do you see it? The difference is constant! No matter how many fives are in the middle, the subtraction $10T_r - T_{r+1}$ always yields $13$.

This is the 'telescoping' effect. In our sum:

$$9S = -T_0 + (10T_0 - T_1) + (10T_1 - T_2) + \dots + (10T_{97} - T_{98}) + 10T_{98}$$

Every bracketed term becomes $13$. Since there are $98$ such brackets (from $r=0$ to $r=97$), the sum of these brackets is $98 	imes 13 = 1274$.

### The Final Assembly

Now we are left with $9S = -T_0 + 1274 + 10T_{98}$. We know $T_0 = 77$, so:

$$9S = -77 + 1274 + 10T_{98} = 1197 + 10T_{98}$$

The problem asks for an expression involving $T_{99}$. We use our previous logic again: $10T_{98} - T_{99} = 13$, which means $10T_{98} = T_{99} + 13$.

Substituting this back, we get:

$$9S = 1197 + T_{99} + 13 = T_{99} + 1210$$

Thus, the sum is:

$$S = \frac{T_{99} + 1210}{9}$$

By comparing this to the form given in the question, we identify $m = 1210$ and $n = 9$. The final result is **$m + n = 1219$**.

Enter Numerical Value:

Visualized Solution

Define the Terms $T_{r}$

The $10 S$ Method

Subtracting $S$ from $10 S$

Evaluating $(10 T_{r} - T_{r + 1})$

Counting the $13$ s

Simplifying the Equation

Expressing $10 T_{98}$ via $T_{99}$

Final Expression for $S$

Finding $m + n$

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The $10 S$ Strategy

The Telescoping Magic

The Final Assembly

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Enter Numerical Value:

Visualized Solution

Define the Terms .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tr​

The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 10S Method

Subtracting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 10S

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (10Tr​−Tr+1​)

Counting the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 13s

Simplifying the Equation

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 10T98​ via .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T99​

Final Expression for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m+n