Problem 14 - Calculus

Question

Simplify the sums \(\sum_{j=1}^{n}\left(f_{j}-f_{j-1}\right)\) and \(\sum_{j=3}^{12}\left(f_{j+1}-f_{j}\right)\).

Step-by-Step Solution

Answer

\(f_n - f_0 \) and \(f_{13} - f_3 \)

Step 1: Understanding the Summation

The first sum is \( \sum_{j=1}^{n}(f_{j} - f_{j-1}) \). This sum type is typical in mathematics where you extend a consecutive cancellation pattern (telescoping sum), resulting only in terms not having a counterpart to cancel with.

Step 2: Apply Telescoping in First Sum

The terms \( f_{1} - f_{0}, f_{2} - f_{1}, \,\ldots\, f_{n} - f_{n-1} \) lead to cancellations: \( f_1 \) cancels with \(-f_1\), \( f_2 \) cancels with \(-f_2\), etc., leaving only \( f_n - f_0 \). Thus, \( \sum_{j=1}^{n}(f_{j} - f_{j-1}) = f_n - f_0 \).

Step 3: Understanding the Second Summation

For the second sum, \( \sum_{j=3}^{12}(f_{j+1} - f_j) \), the cancellation pattern is similar to the telescoping sum pattern found in the first sum.

Step 4: Apply Telescoping in Second Sum

The terms \( f_4 - f_3, f_5 - f_4, \,\ldots\, f_{13} - f_{12} \) similarly cancel: \( f_4 \) cancels with \(-f_4\), \( f_5 \) cancels with \(-f_5\), etc., leaving only \( f_{13} - f_3 \). Thus, \( \sum_{j=3}^{12}(f_{j+1} - f_j) = f_{13} - f_3 \).

Key Concepts

Summation

Summation is a mathematical operation that represents the addition of a sequence of numbers. In the context of the exercises, summation simplifies to look at an entire range of terms rather than individual additions. The summation notation, represented by the Greek letter sigma \(\sum\), allows us to denote these repetitive additions in a compact form. For example, \( \sum_{j=1}^{n} (f_j - f_{j-1}) \) implies that we add such expressions from \(j=1\) to \(j=n\).
Understanding how these sums operate is crucial when you explore further mathematics, as it saves you from writing out long addition sequences and highlights underlying patterns.

Cancellations

Cancellations within a sum occur when opposite terms nullify each other. This concept is particularly useful in simplifying expressions with telescoping series. A telescoping series is a sequence where most terms cancel each other except for a few.
This is clearly seen in the expression \( \sum_{j=1}^{n}(f_{j} - f_{j-1}) \). Here, each \(f_j\) after \(j=1\) has a counterpart \,\(-f_j\), ensuring they cancel out entirely like dominoes. You are left only with \(f_n - f_0\) because \(f_n\) and \(f_0\) have no counterparts to cancel them.
Such cancellations simplify your calculations significantly, converting what seems like a complicated sum into just a subtraction between two edge terms.

Mathematical Patterns

Recognizing mathematical patterns, especially in series and summations, can drastically simplify problem-solving processes. In the exercise, both sums illustrate the use of a telescoping pattern, which is a repeating structure where sequential terms tend to cancel each other.
By identifying this pattern in \( \sum_{j=3}^{12}(f_{j+1} - f_j) \), we realize the sum boils down to \(f_{13} - f_3\).
Mathematical patterns often save time and reduce errors, as you focus only on the start and end terms instead of dealing with each term individually. Knowing these patterns, you can tackle various mathematical challenges with confidence and efficiency.