Math / Algebra

QuestionWrite the sum using sigma notation: $7+9+11+13+\ldots+23=\sum_{n=1}^{A} B$ , where $A=$ $\square$ $B=$ $\square$

Studdy Solution

STEP 1

What is this asking? We need to rewrite the sum $7+9+11+13+\ldots+23$ using sigma notation, which is a compact way of writing sums, and find the values of $A$ and $B$ . Watch out! It's easy to mess up the starting value of $n$ if you're not careful!
Also, make sure the formula inside the sigma accurately generates each term in the sum.

STEP 2

1. Find the pattern
2. Determine the formula
3. Find the upper limit
4. Write in sigma notation

STEP 3

Let's look closely at the given sum: $7 + 9 + 11 + 13 + \ldots + 23$ .
Notice that each term is 2 greater than the previous term.
This means we're dealing with an arithmetic sequence!

STEP 4

The general formula for an arithmetic sequence is $a_n = a_1 + (n-1)d$ , where $a_n$ is the $n$ th term, $a_1$ is the first term, and $d$ is the common difference.

STEP 5

In our case, the first term is $a_1 = 7$ , and the common difference is $d = 2$ .
So, our formula becomes $a_n = 7 + (n-1) \cdot 2$ .

STEP 6

Let's simplify this formula: $a_n = 7 + 2n - 2 = 2n + 5$ .
This formula will give us the $n$ th term in our sequence.

STEP 7

We need to figure out what value of $n$ corresponds to the last term in our sum, which is $23$ .
We can set our formula equal to $23$ and solve for $n$ : $2n + 5 = 23$ .

STEP 8

Subtract 5 from both sides: $2n = 23 - 5 = 18$ .

STEP 9

Divide both sides by 2: $n = \frac{18}{2} = 9$ .
So, there are 9 terms in our sum, and the upper limit of our sigma notation will be $9$ .

STEP 10

Now we can put it all together!
We use the sigma notation $\sum_{n=1}^{A} B$ , where $A$ is the upper limit of the sum and $B$ is the formula for the terms.

STEP 11

We found that $A = 9$ and $B = 2n + 5$ .
Therefore, our sum in sigma notation is $\sum_{n=1}^{9} (2n + 5)$ .

STEP 12

$A = 9$ and $B = 2n + 5$ .
The sum in sigma notation is $\sum_{n=1}^{9} (2n + 5)$ .

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QuestionWrite the sum using sigma notation: $7+9+11+13+\ldots+23=\sum_{n=1}^{A} B$ , where $A=$ $\square$ $B=$ $\square$

Studdy Solution

STEP 1

STEP 2

1. Find the pattern
2. Determine the formula
3. Find the upper limit
4. Write in sigma notation

STEP 3

Let's look closely at the given sum: $7 + 9 + 11 + 13 + \ldots + 23$ .
Notice that each term is 2 greater than the previous term.
This means we're dealing with an arithmetic sequence!

STEP 4

The general formula for an arithmetic sequence is $a_n = a_1 + (n-1)d$ , where $a_n$ is the $n$ th term, $a_1$ is the first term, and $d$ is the common difference.

STEP 5

In our case, the first term is $a_1 = 7$ , and the common difference is $d = 2$ .
So, our formula becomes $a_n = 7 + (n-1) \cdot 2$ .

STEP 6

Let's simplify this formula: $a_n = 7 + 2n - 2 = 2n + 5$ .
This formula will give us the $n$ th term in our sequence.

STEP 7

We need to figure out what value of $n$ corresponds to the last term in our sum, which is $23$ .
We can set our formula equal to $23$ and solve for $n$ : $2n + 5 = 23$ .

STEP 8

Subtract 5 from both sides: $2n = 23 - 5 = 18$ .

STEP 9

Divide both sides by 2: $n = \frac{18}{2} = 9$ .
So, there are 9 terms in our sum, and the upper limit of our sigma notation will be $9$ .

STEP 10

Now we can put it all together!
We use the sigma notation $\sum_{n=1}^{A} B$ , where $A$ is the upper limit of the sum and $B$ is the formula for the terms.

STEP 11

We found that $A = 9$ and $B = 2n + 5$ .
Therefore, our sum in sigma notation is $\sum_{n=1}^{9} (2n + 5)$ .

STEP 12

$A = 9$ and $B = 2n + 5$ .
The sum in sigma notation is $\sum_{n=1}^{9} (2n + 5)$ .

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QuestionWrite the sum using sigma notation: 7+9+11+13+…+23=∑n=1AB7+9+11+13+\ldots+23=\sum_{n=1}^{A} B7+9+11+13+…+23=∑n=1A​B, where A=A=A= □\square□ B=B=B= □\square□

Studdy Solution

STEP 1

STEP 2

1. Find the pattern2. Determine the formula3. Find the upper limit4. Write in sigma notation

STEP 3

Let's look closely at the given sum: 7+9+11+13+…+237 + 9 + 11 + 13 + \ldots + 237+9+11+13+…+23.Notice that each term is **2 greater** than the previous term.This means we're dealing with an **arithmetic sequence**!

STEP 4

The **general formula** for an arithmetic sequence is an=a1+(n−1)da_n = a_1 + (n-1)dan​=a1​+(n−1)d, where ana_nan​ is the nnnth term, a1a_1a1​ is the **first term**, and ddd is the **common difference**.

STEP 5

In our case, the **first term** is a1=7a_1 = 7a1​=7, and the **common difference** is d=2d = 2d=2.So, our formula becomes an=7+(n−1)⋅2a_n = 7 + (n-1) \cdot 2an​=7+(n−1)⋅2.

STEP 6

Let's **simplify** this formula: an=7+2n−2=2n+5a_n = 7 + 2n - 2 = 2n + 5an​=7+2n−2=2n+5.This formula will give us the nnnth term in our sequence.

STEP 7

We need to figure out what value of nnn corresponds to the **last term** in our sum, which is 232323.We can set our formula equal to 232323 and solve for nnn: 2n+5=232n + 5 = 232n+5=23.

STEP 8

**Subtract 5** from both sides: 2n=23−5=182n = 23 - 5 = 182n=23−5=18.

STEP 9

**Divide both sides by 2**: n=182=9n = \frac{18}{2} = 9n=218​=9.So, there are **9 terms** in our sum, and the upper limit of our sigma notation will be 999.

STEP 10

Now we can put it all together!We use the **sigma notation** ∑n=1AB\sum_{n=1}^{A} B∑n=1A​B, where AAA is the **upper limit** of the sum and BBB is the **formula** for the terms.

STEP 11

We found that A=9A = 9A=9 and B=2n+5B = 2n + 5B=2n+5.Therefore, our sum in sigma notation is ∑n=19(2n+5)\sum_{n=1}^{9} (2n + 5)∑n=19​(2n+5).

STEP 12

A=9A = 9A=9 and B=2n+5B = 2n + 5B=2n+5.The sum in sigma notation is ∑n=19(2n+5)\sum_{n=1}^{9} (2n + 5)∑n=19​(2n+5).

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QuestionWrite the sum using sigma notation: $7+9+11+13+\ldots+23=\sum_{n=1}^{A} B$ , where $A=$ $\square$ $B=$ $\square$

1. Find the pattern
2. Determine the formula
3. Find the upper limit
4. Write in sigma notation

Let's look closely at the given sum: $7 + 9 + 11 + 13 + \ldots + 23$ .
Notice that each term is 2 greater than the previous term.
This means we're dealing with an arithmetic sequence!

The general formula for an arithmetic sequence is $a_n = a_1 + (n-1)d$ , where $a_n$ is the $n$ th term, $a_1$ is the first term, and $d$ is the common difference.

In our case, the first term is $a_1 = 7$ , and the common difference is $d = 2$ .
So, our formula becomes $a_n = 7 + (n-1) \cdot 2$ .

Let's simplify this formula: $a_n = 7 + 2n - 2 = 2n + 5$ .
This formula will give us the $n$ th term in our sequence.

We need to figure out what value of $n$ corresponds to the last term in our sum, which is $23$ .
We can set our formula equal to $23$ and solve for $n$ : $2n + 5 = 23$ .

Subtract 5 from both sides: $2n = 23 - 5 = 18$ .

Divide both sides by 2: $n = \frac{18}{2} = 9$ .
So, there are 9 terms in our sum, and the upper limit of our sigma notation will be $9$ .

Now we can put it all together!
We use the sigma notation $\sum_{n=1}^{A} B$ , where $A$ is the upper limit of the sum and $B$ is the formula for the terms.

We found that $A = 9$ and $B = 2n + 5$ .
Therefore, our sum in sigma notation is $\sum_{n=1}^{9} (2n + 5)$ .

$A = 9$ and $B = 2n + 5$ .
The sum in sigma notation is $\sum_{n=1}^{9} (2n + 5)$ .