Unraveling The Summation S_n = Σ[r=1 To N] R(r+1)(1+2^r)

Introduction

In the realm of mathematical series and summations, there exist intriguing problems that challenge our understanding of sequences and their cumulative behavior. One such problem involves the summation S_n = Σ[r=1 to n] u_r, where the general term u_r is defined as r(r+1)(1+2^r). This particular summation presents a unique challenge due to the presence of both polynomial and exponential terms within the summand. To effectively tackle this problem, we need to employ a combination of algebraic manipulation, summation techniques, and a keen eye for patterns. This article will delve into the step-by-step process of unraveling this summation, providing a comprehensive explanation of the methods used and the underlying mathematical principles. Understanding this summation not only enhances our problem-solving skills but also deepens our appreciation for the elegance and interconnectedness of mathematical concepts. By breaking down the problem into manageable parts and applying appropriate techniques, we can arrive at a closed-form expression for S_n, revealing the underlying structure of this intriguing summation.

Understanding the Problem

Before diving into the solution, it's crucial to fully understand the problem statement. We are given the summation S_n = Σ[r=1 to n] u_r, where the term u_r is defined as r(r+1)(1+2^r). Our goal is to find a closed-form expression for S_n, meaning an expression that directly calculates the sum without requiring us to compute each term individually. The presence of both polynomial terms (r(r+1)) and an exponential term (2^r) makes this summation non-trivial. To begin, we can expand the term u_r to gain a better understanding of its structure. This expansion will allow us to separate the polynomial and exponential components, which can then be addressed using different summation techniques. By carefully analyzing the expanded form, we can identify potential patterns or simplifications that will aid in finding a closed-form expression for S_n. This initial understanding and manipulation of the problem are essential steps in formulating a strategy for solving the summation. The key lies in recognizing the interplay between the polynomial and exponential terms and devising a method to handle their combined contribution to the overall sum.

Expanding the Summand

The first step in tackling this summation is to expand the general term u_r. We have u_r = r(r+1)(1+2^r). Expanding this expression, we get:

u_r = r(r+1) + r(r+1)2^r

u_r = r^2 + r + (r^2 + r)2^r

This expansion separates the term into a polynomial part (r^2 + r) and a part that combines a polynomial with an exponential ( (r^2 + r)2^r). This separation is crucial because it allows us to address each part separately using different summation techniques. The polynomial part can be handled using standard summation formulas for sums of powers of integers, while the exponential part requires a more nuanced approach, often involving summation by parts or other specialized methods. By expanding the summand, we have transformed the original problem into a sum of two simpler, more manageable summations. This is a common strategy in dealing with complex summations – breaking them down into smaller, more familiar components. The next step will involve applying appropriate summation techniques to each of these components to find their respective sums.

Separating the Summation

Now that we have expanded the summand, we can separate the original summation into two distinct summations. Recall that S_n = Σ[r=1 to n] u_r, and we have expanded u_r as r^2 + r + (r^2 + r)2^r. Therefore, we can write S_n as:

S_n = Σ[r=1 to n] (r^2 + r) + Σ[r=1 to n] ((r^2 + r)2^r)

Let's denote the first summation as S_1 and the second summation as S_2. Thus,

S_1 = Σ[r=1 to n] (r^2 + r)

S_2 = Σ[r=1 to n] ((r^2 + r)2^r)

This separation allows us to tackle each summation independently. S_1 involves the sum of polynomial terms, which can be readily evaluated using known summation formulas. S_2, on the other hand, presents a greater challenge due to the presence of the exponential term 2^r. This separation is a key step in simplifying the problem, as it allows us to focus our efforts on each part individually. The strategy now is to find closed-form expressions for both S_1 and S_2, and then add them together to obtain the final expression for S_n. The next sections will focus on evaluating each of these summations in detail.

Evaluating S_1

The first summation, S_1 = Σ[r=1 to n] (r^2 + r), involves the sum of polynomial terms. We can further break this down into two separate summations:

S_1 = Σ[r=1 to n] r^2 + Σ[r=1 to n] r

We have well-known formulas for these summations:

Σ[r=1 to n] r = n(n+1)/2

Σ[r=1 to n] r^2 = n(n+1)(2n+1)/6

Substituting these formulas into the expression for S_1, we get:

S_1 = n(n+1)(2n+1)/6 + n(n+1)/2

Now, we can simplify this expression by finding a common denominator and combining the terms:

S_1 = [n(n+1)(2n+1) + 3n(n+1)] / 6

S_1 = n(n+1)(2n+1 + 3) / 6

S_1 = n(n+1)(2n+4) / 6

S_1 = n(n+1)(n+2) / 3

Therefore, we have found a closed-form expression for S_1. This part of the problem was relatively straightforward, utilizing standard summation formulas. The next challenge lies in evaluating S_2, which involves the more complex summation with the exponential term. The evaluation of S_1 serves as a solid foundation, allowing us to focus our attention on the more intricate aspects of S_2. The techniques required for S_2 are more advanced, often involving summation by parts or other clever manipulations.

Tackling S_2: Summation by Parts

Now we turn our attention to the second summation, S_2 = Σ[r=1 to n] ((r^2 + r)2^r). This summation is more challenging due to the presence of both polynomial and exponential terms. A common technique for dealing with such summations is summation by parts, which is the discrete analogue of integration by parts. The formula for summation by parts is:

Σ[r=1 to n] u_r v_(r+1) - v_r = u_n v_(n+1) - u_1 v_1 - Σ[r=1 to n-1] v_(r+1) (u_(r+1) - u_r)

To apply this formula, we need to choose appropriate expressions for u_r and v_r. A strategic choice can significantly simplify the summation. In this case, let's choose:

u_r = r^2 + r

v_r = 2^r

Then, the difference v_(r+1) - v_r is:

v_(r+1) - v_r = 2^(r+1) - 2^r = 2^r(2 - 1) = 2^r

And the difference u_(r+1) - u_r is:

u_(r+1) - u_r = ((r+1)^2 + (r+1)) - (r^2 + r) = (r^2 + 2r + 1 + r + 1) - (r^2 + r) = 2r + 2 = 2(r+1)

Now, we can substitute these expressions into the summation by parts formula. This substitution will transform the original summation into a new form, hopefully one that is easier to evaluate. The key to successful application of summation by parts lies in choosing u_r and v_r such that the resulting summation is simpler than the original. This often involves selecting v_r to be the exponential term and u_r to be the polynomial term, as this tends to reduce the degree of the polynomial in the resulting summation.

Applying Summation by Parts

Substituting our choices for u_r and v_r into the summation by parts formula, we get:

Σ[r=1 to n] ((r^2 + r)2^(r+1) - 2^r) = (n^2 + n)2^(n+1) - (1^2 + 1)2^1 - Σ[r=1 to n-1] 2^(r+1) (2(r+1))

This simplifies to:

Σ[r=1 to n] (r^2 + r)2^r = (n^2 + n)2^(n+1) - 4 - 2Σ[r=1 to n-1] (r+1)2^(r+1)

Now, let's make a substitution k = r + 1, so the summation becomes:

Σ[r=1 to n-1] (r+1)2^(r+1) = Σ[k=2 to n] k2^k

We need to evaluate this new summation, which still involves a polynomial and an exponential term. We can apply summation by parts again. This time, let:

u_k = k

v_k = 2^k

Then, v_(k+1) - v_k = 2^(k+1) - 2^k = 2^k

And u_(k+1) - u_k = (k+1) - k = 1

Applying summation by parts once more, we can further simplify the expression. This iterative application of summation by parts is a common strategy for dealing with summations involving products of polynomials and exponentials. Each application reduces the degree of the polynomial term, eventually leading to a summation that can be evaluated directly. The key is to carefully track the terms and boundary conditions at each step.

Second Application of Summation by Parts

Applying summation by parts again to Σ[k=2 to n] k2^k, we get:

Σ[k=2 to n] k2^k = n2^(n+1) - 2*2^2 - Σ[k=2 to n-1] 2^(k+1)

The remaining summation is a geometric series, which we can easily evaluate:

Σ[k=2 to n-1] 2^(k+1) = 2³⁽²(n-2) - 1) / (2 - 1) = 2³⁽²(n-2) - 1) = 2^(n+1) - 8

Substituting this back into the expression, we get:

Σ[k=2 to n] k2^k = n2^(n+1) - 8 - (2^(n+1) - 8) = n2^(n+1) - 2^(n+1)

Now, we can substitute this result back into the expression for S_2:

S_2 = (n^2 + n)2^(n+1) - 4 - 2(n2^(n+1) - 2^(n+1))

Simplifying this expression will give us a closed-form expression for S_2. This process highlights the power of summation by parts in dealing with complex summations. By applying the technique iteratively, we were able to reduce the problem to a simpler geometric series, which could be easily evaluated.

Simplifying S_2 and Finding S_n

Let's simplify the expression for S_2:

S_2 = (n^2 + n)2^(n+1) - 4 - 2n2^(n+1) + 2*2^(n+1)

S_2 = n^2*2(n+1) + n2^(n+1) - 2n*2^(n+1) + 2^(n+2) - 4

S_2 = *n^2*2(n+1) - n2^(n+1) + 2^(n+2) - 4

Now that we have closed-form expressions for both S_1 and S_2, we can find the closed-form expression for S_n:

S_n = S_1 + S_2

S_n = n(n+1)(n+2) / 3 + *n^2*2(n+1) - n2^(n+1) + 2^(n+2) - 4

This is the final closed-form expression for the summation S_n. This result demonstrates the power of combining algebraic manipulation, summation techniques, and strategic problem-solving to tackle complex mathematical problems. The journey from the initial problem statement to this final expression involved several key steps, including expanding the summand, separating the summation, applying summation by parts, and simplifying the resulting expressions. Each step built upon the previous one, ultimately leading to a solution that reveals the underlying structure of the summation.

Conclusion

In conclusion, we have successfully unraveled the summation S_n = Σ[r=1 to n] r(r+1)(1+2^r) and found a closed-form expression for it. The process involved expanding the summand, separating the summation into simpler parts, applying summation by parts iteratively, and simplifying the resulting expressions. This problem highlights the importance of strategic problem-solving in mathematics, where choosing the right techniques and applying them in the correct order can lead to elegant solutions. The final expression for S_n provides a concise way to calculate the sum for any value of n, avoiding the need to compute each term individually. This journey through the summation problem has not only provided us with a specific solution but also reinforced our understanding of important mathematical concepts and techniques, such as summation by parts and the manipulation of series. The ability to tackle such problems is a testament to the power of mathematical reasoning and the beauty of mathematical structures.