Rewriting And Evaluating Sums Using Summation Properties And Formulas

Mr. Loba Loba · · 6 min read

Table Of Content

  1. Understanding Summation Notation
  2. The Power of Summation Properties
  3. Essential Summation Formulas
  4. Rewriting the Sum
  5. Evaluating the Sum
  6. Breaking Down the Solution
    1. Step 1: Apply the Constant Multiple Rule
    2. Step 2: Use the Sum/Difference Rule
    3. Step 3: Another Constant Multiple Rule
    4. Step 4: Bust Out the Summation Formulas
    5. Step 5: Plug and Chug (with a Calculator, if Needed!)
    6. Step 6: Substitute Back into the Expression
    7. Step 7: Simplify, Simplify, Simplify!
    8. Step 8: Final Multiplication
  7. Real-World Applications

Hey guys! Ever stumbled upon a seemingly monstrous summation problem and felt a shiver down your spine? Fear not! We're about to break down how to tackle these beasts using the power of summation properties and formulas. Let's dive into a specific example and learn how to rewrite and evaluate sums like pros.

Understanding Summation Notation

Before we get our hands dirty with calculations, let’s quickly recap what summation notation is all about. The Greek capital letter sigma (∑\sum) is the star of the show, indicating that we're summing up a series of terms. The expression below the sigma (like k=1k=1) tells us where our index variable (in this case, k) starts, and the expression above the sigma (like 35) tells us where it ends. The expression to the right of the sigma is the formula for the terms we're adding up. So, a summation looks like this:

∑k=13570(k2−2k+1)\sum_{k=1}^{35} 70(k^2 - 2k + 1)

This notation tells us to plug in values for k starting from 1 up to 35, evaluate the expression 70(k2−2k+1)70(k^2 - 2k + 1) for each k, and then add all those results together. Sounds like a lot of work, right? That's where summation properties and formulas come to the rescue!

The Power of Summation Properties

Summation properties are like magic wands that simplify complex sums into manageable pieces. Here are a couple of key properties we'll use:

  1. Constant Multiple Rule: ∑k=1nc⋅ak=c⋅∑k=1nak\sum_{k=1}^{n} c \cdot a_k = c \cdot \sum_{k=1}^{n} a_k
    This rule lets us pull a constant factor out of the summation. Think of it as factoring out a common multiplier to make things easier.
  2. Sum/Difference Rule: ∑k=1n(ak±bk)=∑k=1nak±∑k=1nbk\sum_{k=1}^{n} (a_k \pm b_k) = \sum_{k=1}^{n} a_k \pm \sum_{k=1}^{n} b_k This rule allows us to break up a sum of terms into separate summations. It's like splitting a complicated task into smaller, easier-to-handle subtasks.

Essential Summation Formulas

Besides properties, we have some handy formulas for specific types of sums. These formulas provide shortcuts for calculating sums of common sequences. Here are a few crucial ones:

  1. Sum of the first n natural numbers: ∑k=1nk=n(n+1)2\sum_{k=1}^{n} k = \frac{n(n+1)}{2} This formula gives us a quick way to add up all the integers from 1 to n.
  2. Sum of the squares of the first n natural numbers: ∑k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} This one helps us sum the squares of integers from 1 to n.
  3. Sum of a constant: ∑k=1nc=nc\sum_{k=1}^{n} c = nc This simple formula tells us that adding the same constant c a total of n times is just nc.

With these properties and formulas in our toolbox, we're ready to tackle our original problem!

Rewriting the Sum

Let's revisit our summation:

∑k=13570(k2−2k+1)\sum_{k=1}^{35} 70(k^2 - 2k + 1)

The first thing we can do is apply the Constant Multiple Rule to pull the 70 out of the summation:

70∑k=135(k2−2k+1)70 \sum_{k=1}^{35} (k^2 - 2k + 1)

Next, we can use the Sum/Difference Rule to split the summation into three separate summations:

70[∑k=135k2−∑k=1352k+∑k=1351]70 \left[ \sum_{k=1}^{35} k^2 - \sum_{k=1}^{35} 2k + \sum_{k=1}^{35} 1 \right]

Now, let's apply the Constant Multiple Rule again to the second summation, pulling out the 2:

70[∑k=135k2−2∑k=135k+∑k=1351]70 \left[ \sum_{k=1}^{35} k^2 - 2 \sum_{k=1}^{35} k + \sum_{k=1}^{35} 1 \right]

We've successfully rewritten the sum into a form where we can apply our summation formulas directly!

Evaluating the Sum

Now comes the fun part – plugging in the formulas and crunching the numbers. We have three summations to evaluate:

  1. ∑k=135k2\sum_{k=1}^{35} k^2 : This is the sum of squares, so we use the formula ∑k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}. Plugging in n = 35, we get:

    35(35+1)(2(35)+1)6=35⋅36⋅716=14910\frac{35(35+1)(2(35)+1)}{6} = \frac{35 \cdot 36 \cdot 71}{6} = 14910

  2. ∑k=135k\sum_{k=1}^{35} k : This is the sum of the first n natural numbers, so we use the formula ∑k=1nk=n(n+1)2\sum_{k=1}^{n} k = \frac{n(n+1)}{2}. Plugging in n = 35, we get:

    35(35+1)2=35⋅362=630\frac{35(35+1)}{2} = \frac{35 \cdot 36}{2} = 630

  3. ∑k=1351\sum_{k=1}^{35} 1 : This is the sum of a constant, so we use the formula ∑k=1nc=nc\sum_{k=1}^{n} c = nc. Here, c = 1 and n = 35, so the sum is simply:

    35⋅1=3535 \cdot 1 = 35

Now, let's substitute these values back into our expression:

70[14910−2(630)+35]70 [14910 - 2(630) + 35]

Simplify the expression inside the brackets:

70[14910−1260+35]=70[13685]70 [14910 - 1260 + 35] = 70 [13685]

Finally, multiply by 70:

70⋅13685=95795070 \cdot 13685 = 957950

So, the value of the original summation is 957,950. That's a big number, but we conquered it using our summation superpowers!

Breaking Down the Solution

Okay, let’s break down the solution step-by-step to really nail this concept. Remember our original summation?

∑k=13570(k2−2k+1)\sum_{k=1}^{35} 70(k^2 - 2k + 1)

Step 1: Apply the Constant Multiple Rule

The first smart move is to pull that 70 out front. It’s like saying, “Hey, 70, you just hang out here while we deal with the rest.” This simplifies things:

70∑k=135(k2−2k+1)70 \sum_{k=1}^{35} (k^2 - 2k + 1)

Step 2: Use the Sum/Difference Rule

Next up, we split the summation into smaller, bite-sized pieces. Think of it as dividing a pizza so everyone gets their fair share. We’re essentially distributing the summation across each term:

70[∑k=135k2−∑k=1352k+∑k=1351]70 \left[ \sum_{k=1}^{35} k^2 - \sum_{k=1}^{35} 2k + \sum_{k=1}^{35} 1 \right]

Step 3: Another Constant Multiple Rule

See that 2 chilling inside the second summation? Let’s kick it out front too. Constants love hanging outside summations, it seems:

70[∑k=135k2−2∑k=135k+∑k=1351]70 \left[ \sum_{k=1}^{35} k^2 - 2 \sum_{k=1}^{35} k + \sum_{k=1}^{35} 1 \right]

Step 4: Bust Out the Summation Formulas

Now for the magic! We’ve got three summations, and each one has a special formula waiting to be used:

  • ∑k=135k2\sum_{k=1}^{35} k^2: This is the sum of squares. We’ll use n(n+1)(2n+1)6\frac{n(n+1)(2n+1)}{6} with n = 35.
  • ∑k=135k\sum_{k=1}^{35} k: This is the sum of the first n natural numbers. We’ll use n(n+1)2\frac{n(n+1)}{2} with n = 35.
  • ∑k=1351\sum_{k=1}^{35} 1: This is just adding 1 to itself 35 times. Easy peasy!

Step 5: Plug and Chug (with a Calculator, if Needed!)

Let’s plug in the values and see what we get:

  • Sum of squares: 35(35+1)(2(35)+1)6=14910\frac{35(35+1)(2(35)+1)}{6} = 14910
  • Sum of natural numbers: 35(35+1)2=630\frac{35(35+1)}{2} = 630
  • Sum of 1s: 35

Step 6: Substitute Back into the Expression

Now we swap out those summations with their calculated values:

70[14910−2(630)+35]70 [14910 - 2(630) + 35]

Step 7: Simplify, Simplify, Simplify!

Time to do some arithmetic:

70[14910−1260+35]=70[13685]70 [14910 - 1260 + 35] = 70 [13685]

Step 8: Final Multiplication

One last step! Multiply by that 70 we pulled out at the beginning:

70⋅13685=95795070 \cdot 13685 = 957950

Real-World Applications

You might be thinking,

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