Summation Calculations: Solve Series & Sequences Now!

Hey guys! Let's dive into the fascinating world of summation calculations. Summation, also known as sigma notation, is a concise and powerful way to represent the sum of a series of terms. Mastering summation is crucial for various fields, including calculus, statistics, and computer science. In this article, we'll tackle a series of summation problems, breaking down each step to ensure you understand the process thoroughly. So, grab your calculators, and let's get started!

Understanding Summation Notation

Before we jump into solving specific problems, let's quickly recap the basics of summation notation. The Greek letter sigma (Σ) is used to denote summation. The general form looks like this:

Σ[expression] from i = start to end

Where:

• Σ is the summation symbol.
• expression is the formula or function you're summing.
• i is the index variable.
• start is the starting value of the index.
• end is the ending value of the index.

For example, Σ i from i=1 to 5 means we need to sum the values of i as i ranges from 1 to 5. This would be 1 + 2 + 3 + 4 + 5 = 15. Knowing this foundation, we can handle more complex summations.

Solving Summation Problems: A Step-by-Step Approach

Now, let's tackle the given summation problems one by one. We'll break down each problem, showing you how to compute the sums systematically. Remember, the key is to substitute each value of the index variable into the expression and then add up the results.

a) Σ -2(2i – 3) from i=-3 to 4

In this problem, we need to calculate the sum of -2(2i – 3) as i varies from -3 to 4. Let's break it down:

1. Substitute each value of i from -3 to 4 into the expression -2(2i – 3):
   • For i = -3: -2(2(-3) – 3) = -2(-6 – 3) = -2(-9) = 18
   • For i = -2: -2(2(-2) – 3) = -2(-4 – 3) = -2(-7) = 14
   • For i = -1: -2(2(-1) – 3) = -2(-2 – 3) = -2(-5) = 10
   • For i = 0: -2(2(0) – 3) = -2(0 – 3) = -2(-3) = 6
   • For i = 1: -2(2(1) – 3) = -2(2 – 3) = -2(-1) = 2
   • For i = 2: -2(2(2) – 3) = -2(4 – 3) = -2(1) = -2
   • For i = 3: -2(2(3) – 3) = -2(6 – 3) = -2(3) = -6
   • For i = 4: -2(2(4) – 3) = -2(8 – 3) = -2(5) = -10
2. Add up all the results: 18 + 14 + 10 + 6 + 2 + (-2) + (-6) + (-10) = 32

Therefore, Σ -2(2i – 3) from i=-3 to 4 equals 32. Understanding how to substitute and sum is fundamental to solving these problems.

b) Σ i² from i= −3 to 2

Here, we need to calculate the sum of the squares of i as i goes from -3 to 2. This involves squaring each value of i and then adding them up:

1. Substitute each value of i from -3 to 2 into the expression i²:
   • For i = -3: (-3)² = 9
   • For i = -2: (-2)² = 4
   • For i = -1: (-1)² = 1
   • For i = 0: (0)² = 0
   • For i = 1: (1)² = 1
   • For i = 2: (2)² = 4
2. Add up all the results: 9 + 4 + 1 + 0 + 1 + 4 = 19

Thus, Σ i² from i= −3 to 2 equals 19. This problem reinforces the importance of correctly applying the exponent before summing the terms.

c) Σ 2i from i=-5 to 5

In this case, we are summing 2i as i ranges from -5 to 5. This is a straightforward calculation involving multiplying each i by 2 and then summing the results.

1. Substitute each value of i from -5 to 5 into the expression 2i:
   • For i = -5: 2(-5) = -10
   • For i = -4: 2(-4) = -8
   • For i = -3: 2(-3) = -6
   • For i = -2: 2(-2) = -4
   • For i = -1: 2(-1) = -2
   • For i = 0: 2(0) = 0
   • For i = 1: 2(1) = 2
   • For i = 2: 2(2) = 4
   • For i = 3: 2(3) = 6
   • For i = 4: 2(4) = 8
   • For i = 5: 2(5) = 10
2. Add up all the results: -10 + (-8) + (-6) + (-4) + (-2) + 0 + 2 + 4 + 6 + 8 + 10 = 0

Therefore, Σ 2i from i=-5 to 5 equals 0. Notice how the positive and negative terms canceled each other out, making the final sum zero. This is a common occurrence in summations involving symmetrical ranges and linear expressions.

d) Σ -3k from k = 1 to k

This question seems a bit ambiguous because the upper limit of the summation is given as k, which is also the index variable. This typically means we're looking for a general formula rather than a specific numerical value. We're summing -3k from k=1 to some arbitrary value k. This is essentially the sum of the first k multiples of -3.

The sum can be written as:

Σ -3k from k = 1 to k = -3(1) + -3(2) + -3(3) + ... + -3(k)

We can factor out the -3:

-3(1 + 2 + 3 + ... + k)

The sum of the first k natural numbers is given by the formula k(k + 1) / 2. Therefore, the summation becomes:

-3 * [k(k + 1) / 2]

So, Σ -3k from k = 1 to k = -3k(k + 1) / 2. This is the general formula for this summation.

e) Σ 3i-2 from i=-6 to 2

Here, we need to sum the expression 3i-2 as i goes from -6 to 2. This involves multiplying each i by 3, subtracting 2, and then summing the results.

1. Substitute each value of i from -6 to 2 into the expression 3i-2:
   • For i = -6: 3(-6) - 2 = -18 - 2 = -20
   • For i = -5: 3(-5) - 2 = -15 - 2 = -17
   • For i = -4: 3(-4) - 2 = -12 - 2 = -14
   • For i = -3: 3(-3) - 2 = -9 - 2 = -11
   • For i = -2: 3(-2) - 2 = -6 - 2 = -8
   • For i = -1: 3(-1) - 2 = -3 - 2 = -5
   • For i = 0: 3(0) - 2 = 0 - 2 = -2
   • For i = 1: 3(1) - 2 = 3 - 2 = 1
   • For i = 2: 3(2) - 2 = 6 - 2 = 4
2. Add up all the results: -20 + (-17) + (-14) + (-11) + (-8) + (-5) + (-2) + 1 + 4 = -72

Thus, Σ 3i-2 from i=-6 to 2 equals -72. Careful arithmetic is crucial for getting the correct answer in these types of summations.

f) Σ i from i= 1 to 1

This is a simple summation where we sum i from i=1 to 1. Since the start and end values are the same, we only have one term to consider.

1. Substitute i = 1 into the expression i:
   • For i = 1: 1
2. The sum is simply 1.

Therefore, Σ i from i= 1 to 1 equals 1. This illustrates a special case of summation where the range consists of a single value.

g) Σ (i+3)/2 from i=-3 to 2

In this problem, we are summing the expression (i+3)/2 as i ranges from -3 to 2. This involves adding 3 to each i, dividing by 2, and then summing the results.

1. Substitute each value of i from -3 to 2 into the expression (i+3)/2:
   • For i = -3: (-3 + 3) / 2 = 0 / 2 = 0
   • For i = -2: (-2 + 3) / 2 = 1 / 2 = 0.5
   • For i = -1: (-1 + 3) / 2 = 2 / 2 = 1
   • For i = 0: (0 + 3) / 2 = 3 / 2 = 1.5
   • For i = 1: (1 + 3) / 2 = 4 / 2 = 2
   • For i = 2: (2 + 3) / 2 = 5 / 2 = 2.5
2. Add up all the results: 0 + 0.5 + 1 + 1.5 + 2 + 2.5 = 7.5

Thus, Σ (i+3)/2 from i=-3 to 2 equals 7.5. Dealing with fractions or decimals in summation requires careful attention to detail.

h) Σ (3i-3) from i= 0 to 2

Here, we need to sum the expression 3i-3 as i goes from 0 to 2. We'll multiply each i by 3, subtract 3, and then sum the results.

1. Substitute each value of i from 0 to 2 into the expression 3i-3:
   • For i = 0: 3(0) - 3 = 0 - 3 = -3
   • For i = 1: 3(1) - 3 = 3 - 3 = 0
   • For i = 2: 3(2) - 3 = 6 - 3 = 3
2. Add up all the results: -3 + 0 + 3 = 0

Therefore, Σ (3i-3) from i= 0 to 2 equals 0. Again, notice how the terms can cancel out, simplifying the final result.

i) Σ 2 from i=-3 to 2

In this problem, we are summing the constant value 2 as i ranges from -3 to 2. This means we are adding 2 to itself a certain number of times.

1. Determine the number of terms in the summation. The range is from -3 to 2, which includes the following values: -3, -2, -1, 0, 1, 2. There are 6 terms in total.
2. Multiply the constant value (2) by the number of terms (6): 2 * 6 = 12

Thus, Σ 2 from i=-3 to 2 equals 12. Summing a constant value is equivalent to multiplying the constant by the number of terms in the range.

Conclusion: Mastering Summation Calculations

Alright guys, we've tackled a variety of summation problems, and hopefully, you now have a solid understanding of how to calculate summations. Remember, the key is to carefully substitute each value of the index variable into the expression and then add up the results. Practice makes perfect, so keep working on different types of summation problems to solidify your skills.

Summation is a powerful tool in mathematics and other fields. By mastering it, you'll be well-equipped to tackle more advanced concepts in calculus, statistics, and beyond. Keep up the great work, and happy summing!