Math Problem Statement

For each of the summations, use the formula for the sum of the first n integers either to evaluate the sum or express it in closed form.

For each of the summations, use the formula for the sum of the first n integers either to evaluate the sum or express it in closed form.

Solution

To solve the given summations, we can use the formula for the sum of an arithmetic series. The general formula for the sum of an arithmetic series is:

Sn=n2×(a+l)S_n = \frac{n}{2} \times (a + l)

where:

  • SnS_n is the sum of the series,
  • nn is the number of terms,
  • aa is the first term, and
  • ll is the last term.

Alternatively, if you know the number of terms nn, the sum can also be expressed as:

Sn=n2×(2a+(n1)×d)S_n = \frac{n}{2} \times (2a + (n-1) \times d)

where dd is the common difference between terms (which is 1 in this case).

Part (a) - Sum from 9 to 600

The arithmetic series starts at 9 and ends at 600. We need to find the sum of all integers from 9 to 600.

  1. First term aa = 9
  2. Last term ll = 600
  3. Common difference dd = 1

To find the number of terms nn, use the formula for the nth term of an arithmetic series:

l=a+(n1)×dl = a + (n - 1) \times d

Substituting the values:

600=9+(n1)×1600 = 9 + (n - 1) \times 1

Solving for nn:

6009=n1600 - 9 = n - 1 n=592n = 592

Now, use the sum formula:

S592=5922×(9+600)=296×609=180,264S_{592} = \frac{592}{2} \times (9 + 600) = 296 \times 609 = 180,264

So, the sum 9+10+11++600=180,2649 + 10 + 11 + \cdots + 600 = 180,264.

Part (b) - Sum from 9 to kk

Here, the arithmetic series starts at 9 and ends at kk. We need to express the sum in a closed form.

  1. First term aa = 9
  2. Last term l=kl = k
  3. Common difference d=1d = 1

The number of terms nn is:

n=k9+1=k8n = k - 9 + 1 = k - 8

Now, the sum can be expressed as:

Sn=n2×(a+l)=k82×(9+k)S_n = \frac{n}{2} \times (a + l) = \frac{k - 8}{2} \times (9 + k)

So, the sum 9+10+11++k=k82×(9+k)9 + 10 + 11 + \cdots + k = \frac{k - 8}{2} \times (9 + k).


If you have any further questions or need additional details, feel free to ask!

Relative Questions:

  1. How would the solution change if the first term was 10 instead of 9?
  2. What if the common difference dd was 2 instead of 1?
  3. Can you derive the sum formula for arithmetic series using induction?
  4. How would you calculate the sum if the last term was a variable nn rather than a constant?
  5. How can you verify the result using another method or formula?

Tip:

Always double-check your formulas and calculations, especially when dealing with large sums or algebraic expressions.

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Math Problem Analysis

Mathematical Concepts

Arithmetic Series
Summation
Algebra

Formulas

Sum of an arithmetic series: S_n = (n/2) * (a + l)
Sum of an arithmetic series (alternative): S_n = (n/2) * (2a + (n-1) * d)

Theorems

Arithmetic Progression (AP) Formula

Suitable Grade Level

Grades 9-12