Evaluate Summation Of -2n-3 From N=2 To 10 A Step By Step Guide

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In this comprehensive guide, we will delve into the step-by-step evaluation of the summation $\sum_{n=2}^{10}-2n-3$. This problem falls under the domain of mathematics, specifically focusing on the concepts of sequences and series. Understanding how to evaluate summations is crucial in various fields, including calculus, discrete mathematics, and computer science. We will break down the problem into manageable parts, ensuring a clear and concise understanding of each step involved. Let's embark on this journey of mathematical exploration and unravel the solution together.

Understanding the Summation Notation

Before we dive into the specifics of this particular problem, let's first establish a solid foundation by understanding the summation notation itself. The Greek capital letter sigma, $\sum$, is the symbol used to denote summation. It essentially represents the process of adding a series of terms together. The general form of a summation is given by:

∑i=mnai\sum_{i=m}^{n} a_i

Where:

  • $\sum$ is the summation symbol.
  • $i$ is the index of summation, which is a variable that takes on integer values.
  • $m$ is the lower limit of summation, the starting value of the index.
  • $n$ is the upper limit of summation, the ending value of the index.
  • $a_i$ is the expression or formula that defines the terms being summed. This expression depends on the index $i$.

In essence, the summation notation tells us to substitute each integer value of $i$ from $m$ to $n$ into the expression $a_i$, and then add all the resulting terms together. For instance,

∑i=14i2=12+22+32+42=1+4+9+16=30\sum_{i=1}^{4} i^2 = 1^2 + 2^2 + 3^2 + 4^2 = 1 + 4 + 9 + 16 = 30

This foundational understanding of summation notation is crucial for tackling more complex problems. It allows us to express concisely what would otherwise be a lengthy addition process. Now that we have a firm grasp of the notation, let's proceed to analyze the specific summation we aim to evaluate.

Breaking Down the Summation $\sum_{n=2}^{10}-2n-3$

Now, let's focus on the summation at hand: $\sum_{n=2}^{10}-2n-3$. Our primary goal is to evaluate this expression, which means we need to find the sum of the terms generated by the expression $-2n-3$ as $n$ ranges from 2 to 10. To do this effectively, we can break down the summation into smaller, more manageable steps. The expression $-2n-3$ represents a linear function of $n$. As $n$ changes, the value of this expression will also change, and we need to sum up all these values within the specified range.

Step 1: Understanding the Index and Limits of Summation

The first step in evaluating any summation is to clearly identify the index of summation and the limits. In this case, the index of summation is $n$, and it ranges from a lower limit of 2 to an upper limit of 10. This means that we will be substituting the integer values 2, 3, 4, ..., 10 into the expression $-2n-3$. There are a total of 9 terms to be summed in this case (10 - 2 + 1 = 9). This understanding of the limits helps us determine the number of terms we need to calculate and add together.

Step 2: Expanding the Summation

To visualize the summation more clearly, we can expand it by writing out the individual terms. This involves substituting each value of $n$ from 2 to 10 into the expression $-2n-3$ and writing them as a sum:

∑n=210−2n−3=(−2(2)−3)+(−2(3)−3)+(−2(4)−3)+⋯+(−2(10)−3)\sum_{n=2}^{10}-2n-3 = (-2(2)-3) + (-2(3)-3) + (-2(4)-3) + \cdots + (-2(10)-3)

This expansion allows us to see the pattern and the individual values that need to be added. It's a crucial step in understanding the summation and preparing for the next phase of calculation.

Step 3: Calculating Individual Terms

Now that we have expanded the summation, we can calculate the value of each term. This involves simple arithmetic operations. For each value of $n$, we substitute it into the expression $-2n-3$ and perform the calculation:

  • For $n=2$: $-2(2) - 3 = -4 - 3 = -7$
  • For $n=3$: $-2(3) - 3 = -6 - 3 = -9$
  • For $n=4$: $-2(4) - 3 = -8 - 3 = -11$
  • For $n=5$: $-2(5) - 3 = -10 - 3 = -13$
  • For $n=6$: $-2(6) - 3 = -12 - 3 = -15$
  • For $n=7$: $-2(7) - 3 = -14 - 3 = -17$
  • For $n=8$: $-2(8) - 3 = -16 - 3 = -19$
  • For $n=9$: $-2(9) - 3 = -18 - 3 = -21$
  • For $n=10$: $-2(10) - 3 = -20 - 3 = -23$

Now we have the individual terms of the summation: -7, -9, -11, -13, -15, -17, -19, -21, and -23. The next step is to add these terms together.

Summing the Terms: Arithmetic Series Approach

After calculating the individual terms, the next logical step is to add them together to find the final sum. We have the following sequence of terms: -7, -9, -11, -13, -15, -17, -19, -21, -23. Observing this sequence, we can recognize that it forms an arithmetic series. An arithmetic series is a sequence of numbers in which the difference between any two consecutive terms is constant. This constant difference is called the common difference.

In our case, the common difference is -2 (e.g., -9 - (-7) = -2, -11 - (-9) = -2, and so on). Recognizing this pattern allows us to use the formula for the sum of an arithmetic series, which provides a more efficient method for calculating the sum compared to adding each term individually.

The formula for the sum of an arithmetic series is:

Sn=n2(a1+an)S_n = \frac{n}{2}(a_1 + a_n)

Where:

  • $S_n$ is the sum of the first $n$ terms of the series.
  • $n$ is the number of terms.
  • $a_1$ is the first term.
  • $a_n$ is the last term.

Applying the Formula

In our summation, we have:

  • $n = 9$ (there are 9 terms from $n=2$ to $n=10$)
  • $a_1 = -7$ (the first term when $n=2$)
  • $a_n = -23$ (the last term when $n=10$)

Substituting these values into the formula, we get:

S9=92(−7+(−23))S_9 = \frac{9}{2}(-7 + (-23))

S9=92(−30)S_9 = \frac{9}{2}(-30)

S9=9×(−15)S_9 = 9 \times (-15)

S9=−135S_9 = -135

Therefore, the sum of the series is -135. This method provides a quick and efficient way to calculate the sum, especially for arithmetic series with a large number of terms.

Alternative Approach: Separating the Summation

Another approach to evaluating the summation $\sum_{n=2}^{10}-2n-3$ involves separating the summation into two simpler summations. This technique utilizes the properties of summation, which allow us to distribute the summation symbol over addition and multiplication by a constant. The key idea is to break down the original summation into components that are easier to evaluate individually.

Using Summation Properties

The summation properties we will use are:

  1. Sum of a constant times a term: $\sum_{i=m}^{n} c \cdot a_i = c \sum_{i=m}^{n} a_i$, where $c$ is a constant.
  2. Sum of a constant: $\sum_{i=m}^{n} c = c(n - m + 1)$, where $c$ is a constant.
  3. Sum of a sum/difference: $\sum_{i=m}^{n} (a_i \pm b_i) = \sum_{i=m}^{n} a_i \pm \sum_{i=m}^{n} b_i$

Applying these properties to our summation, we can rewrite it as follows:

∑n=210−2n−3=∑n=210(−2n)+∑n=210(−3)\sum_{n=2}^{10}-2n-3 = \sum_{n=2}^{10}(-2n) + \sum_{n=2}^{10}(-3)

Now, we can further simplify the first summation by factoring out the constant -2:

∑n=210(−2n)=−2∑n=210n\sum_{n=2}^{10}(-2n) = -2\sum_{n=2}^{10}n

So, our original summation is now broken down into two simpler summations:

∑n=210−2n−3=−2∑n=210n+∑n=210(−3)\sum_{n=2}^{10}-2n-3 = -2\sum_{n=2}^{10}n + \sum_{n=2}^{10}(-3)

Evaluating the Simpler Summations

Now we need to evaluate each of these simpler summations. The first one, $\sum_{n=2}^{10}n$, represents the sum of the integers from 2 to 10. We can use the formula for the sum of an arithmetic series or a variation of the formula for the sum of the first $n$ integers. The formula for the sum of the first $n$ integers is:

∑i=1ni=n(n+1)2\sum_{i=1}^{n} i = \frac{n(n+1)}{2}

To use this formula, we can calculate the sum of integers from 1 to 10 and then subtract the sum of integers from 1 to 1:

∑n=210n=∑n=110n−∑n=11n\sum_{n=2}^{10}n = \sum_{n=1}^{10}n - \sum_{n=1}^{1}n

∑n=210n=10(10+1)2−1(1+1)2\sum_{n=2}^{10}n = \frac{10(10+1)}{2} - \frac{1(1+1)}{2}

∑n=210n=10(11)2−1(2)2\sum_{n=2}^{10}n = \frac{10(11)}{2} - \frac{1(2)}{2}

∑n=210n=55−1=54\sum_{n=2}^{10}n = 55 - 1 = 54

Now let's evaluate the second summation, $\sum_{n=2}^{10}(-3)$. This is the sum of a constant (-3) repeated from $n=2$ to $n=10$. We can use the formula for the sum of a constant:

∑i=mnc=c(n−m+1)\sum_{i=m}^{n} c = c(n - m + 1)

In our case, $c = -3$, $m = 2$, and $n = 10$, so:

∑n=210(−3)=−3(10−2+1)=−3(9)=−27\sum_{n=2}^{10}(-3) = -3(10 - 2 + 1) = -3(9) = -27

Combining the Results

Now that we have evaluated both simpler summations, we can combine the results to find the final sum:

∑n=210−2n−3=−2∑n=210n+∑n=210(−3)\sum_{n=2}^{10}-2n-3 = -2\sum_{n=2}^{10}n + \sum_{n=2}^{10}(-3)

∑n=210−2n−3=−2(54)+(−27)\sum_{n=2}^{10}-2n-3 = -2(54) + (-27)

∑n=210−2n−3=−108−27\sum_{n=2}^{10}-2n-3 = -108 - 27

∑n=210−2n−3=−135\sum_{n=2}^{10}-2n-3 = -135

Thus, we arrive at the same result, -135, using this alternative approach. This method demonstrates the flexibility and power of summation properties in simplifying complex expressions.

Conclusion: The Final Evaluation

In conclusion, we have successfully evaluated the summation $\sum_{n=2}^{10}-2n-3$ using two different methods: the arithmetic series approach and the separation of summation approach. Both methods led us to the same final answer of -135. This consistency in the results reinforces the correctness of our calculations and highlights the versatility of mathematical techniques in problem-solving. By breaking down the problem into smaller, manageable steps, we have gained a deeper understanding of the summation process and the underlying mathematical concepts involved. Whether you choose to use the arithmetic series formula or the summation properties, the key is to approach the problem systematically and carefully. This detailed exploration serves as a valuable guide for tackling similar summation problems in the future. The ability to evaluate summations is a fundamental skill in mathematics and has applications in various fields. By mastering these techniques, you can confidently tackle a wide range of mathematical challenges.