Let for n = 1, 2, ..., 50, S_n be the sum of the infinite geometric progression whose first term is

Q: Wait, why do we reject the negative root or why is the sum of G.P. formula $S_n = \frac{a}{1-r}$ valid here?

The formula $S_n = \frac{a}{1-r}$ is valid for any infinite geometric progression where the absolute value of the common ratio $|r| < 1$. In our problem, $r = \frac{1}{(n+1)^2}$. For $n \ge 1$, $(n+1)^2 \ge 4$, so $r \le \frac{1}{4}$, which is clearly less than 1. Thus, the formula is perfectly applicable.

Q: How do I know when to use the telescoping series method?

Whenever you see a summation of the form $\sum (f(n) - f(n+1))$, it is a huge red flag for a telescoping series. In this problem, after simplifying, we got $2(\frac{1}{n+1} - \frac{1}{n+2})$. Notice how the second term of the $n$-th bracket cancels with the first term of the $(n+1)$-th bracket. That is the hallmark of telescoping!

Q: Why did we split $S_n$ into $n^2 + 1 - \frac{2}{n+2}$? Could we have just summed it as is?

You *could* try to sum it as is, but it would be a nightmare of algebraic complexity. By splitting $S_n$ into a polynomial part ($n^2+1$) and a fractional part ($-\frac{2}{n+2}$), we isolate the parts that are easy to sum using standard formulas ($\sum n^2$, $\sum n$) from the part that requires the telescoping technique. It is all about simplifying the problem into manageable chunks.

Question

Let for $n = 1, 2, \dots, 50, S_n$ be the sum of the infinite geometric progression whose first term is $n^2$ and whose common ratio is $\frac{1}{(n+1)^2}$. Then the value of $\frac{1}{26} + \sum_{n=1}^{50} (S_n + \frac{2}{n+1} - n - 1)$ is equal to

Sigma Percentile · Accepted Answer

### The Elegance of Infinite Series: A Journey Through $S_n$

Welcome, fellow traveler of the mathematical landscape. Today, we are going to dismantle a problem that, at first glance, looks like a daunting mountain of algebra.

But as we peel back the layers, you will see that it is actually a beautifully choreographed dance of numbers. We are dealing with an infinite geometric progression, $S_n$, where the first term is $a = n^2$ and the common ratio is $r = \frac{1}{(n+1)^2}$.

Our goal is to evaluate the sum $\frac{1}{26} + \sum_{n=1}^{50} (S_n + \frac{2}{n+1} - n - 1)$. Let us begin by taming the beast, $S_n$.

### Phase 1: Taming the Infinite G.P.

Recall the fundamental formula for the sum of an infinite geometric progression: $S_n = \frac{a}{1-r}$. Substituting our given values, we get:

$$S_n = \frac{n^2}{1 - \frac{1}{(n+1)^2}}$$

Now, do not let the fraction intimidate you. Let us find a common denominator in the denominator:

$$S_n = \frac{n^2}{\frac{(n+1)^2 - 1}{(n+1)^2}}$$

This simplifies beautifully to:

$$S_n = \frac{n^2(n+1)^2}{(n+1)^2 - 1}$$

Expanding the denominator, $(n+1)^2 - 1 = n^2 + 2n + 1 - 1 = n^2 + 2n = n(n+2)$. Thus, $S_n = \frac{n^2(n+1)^2}{n(n+2)}$.

Canceling the $n$, we arrive at the elegant form:

$$S_n = \frac{n(n+1)^2}{n+2}$$

### Phase 2: The Art of Algebraic Decomposition

Now, we need to make this expression ready for summation. We want to separate the polynomial part from the fractional part.

Let us expand the numerator:

$$S_n = \frac{n(n^2+2n+1)}{n+2} = \frac{n^3+2n^2+n}{n+2}$$

By performing a clever bit of polynomial division or rearrangement, we can write $n^3+2n^2$ as $n^2(n+2)$. So:

$$S_n = \frac{n^2(n+2) + n}{n+2} = n^2 + \frac{n}{n+2}$$

To simplify $\frac{n}{n+2}$, we add and subtract 2 in the numerator: $\frac{n+2-2}{n+2} = 1 - \frac{2}{n+2}$. Therefore, our final, simplified form for $S_n$ is:

$$S_n = n^2 + 1 - \frac{2}{n+2}$$

This is the breakthrough we needed!

### Phase 3: The Telescoping Magic

Now, let us look at the expression inside our summation: $S_n + \frac{2}{n+1} - n - 1$. Substituting our new form of $S_n$, we get:

$$(n^2 + 1 - \frac{2}{n+2}) + \frac{2}{n+1} - n - 1$$

Notice the cancellation: the $+1$ and $-1$ vanish, leaving us with $n^2 - n + \frac{2}{n+1} - \frac{2}{n+2}$. We can group this as $(n^2 - n) + 2(\frac{1}{n+1} - \frac{1}{n+2})$.

This is where the magic happens. The first part is a simple polynomial sum, and the second part is a classic telescoping series.

For the polynomial part, $\sum_{n=1}^{50} (n^2 - n) = \sum n^2 - \sum n$. Using the standard formulas:

$$\sum_{n=1}^{50} n^2 = \frac{50 \cdot 51 \cdot 101}{6} = 42925$$

$$\sum_{n=1}^{50} n = \frac{50 \cdot 51}{2} = 1275$$

Their difference is $41650$.

For the telescoping part, $2 \sum_{n=1}^{50} (\frac{1}{n+1} - \frac{1}{n+2})$, we expand the terms:

$$2 \left[ \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \dots + \left(\frac{1}{51} - \frac{1}{52}\right) \right]$$

Every middle term cancels out, leaving only $2(\frac{1}{2} - \frac{1}{52}) = 1 - \frac{1}{26} = \frac{25}{26}$.

### Conclusion: The Final Cancellation

Finally, we add the $\frac{1}{26}$ from the original expression:

$$\frac{1}{26} + 41650 + \left(1 - \frac{1}{26}\right)$$

The $\frac{1}{26}$ and $-\frac{1}{26}$ cancel out perfectly, leaving us with $41650 + 1 = 41651$.

And there you have it—a complex problem solved through the elegance of algebraic decomposition and the beauty of telescoping series. Never fear the complexity; just look for the pattern. The final answer is **41651**.

Enter Numerical Value:

Visualized Solution

Defining the Infinite G.P. Sum $S_{n}$

Simplifying the Expression for $S_{n}$

Algebraic Decomposition of $S_{n}$

Simplifying the Target Expression

Summing the Polynomial Terms

Evaluating the Telescoping Sum

Final Calculation and Result

The Sigma Insight: Sum of Special Series

The Elegance of Infinite Series

Phase 1

Phase 2

Phase 3

Conclusion

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Enter Numerical Value:

Visualized Solution

Defining the Infinite G.P. Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sn​

Simplifying the Expression for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sn​

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