41–48 ■ find the sum.
41. $ \sum_{k=1}^4 k $
42. $ \sum_{k=1}^4 k^2 $
43. $ \sum_{k=1}^3 \frac{1}{k} $
44. $ \sum_{j=1}^{100} (-1)^j $
45. $ \sum_{i=1}^8 \left 1 + (-1)^i \
ight $
46. $ \sum_{i=4}^{12} 10 $
47. $ \sum_{k=1}^5 2^{k-1} $
48. $ \sum_{i=1}^3 i2^i $

Question

Accepted Answer

### Problem 41: $\sum_{k=1}^{4} k$
# Explanation:
## Step1: Expand the summation
Substitute $ k = 1, 2, 3, 4 $ into the term $ k $:
$ \sum_{k=1}^{4} k = 1 + 2 + 3 + 4 $
## Step2: Calculate the sum
Add the numbers together:
$ 1 + 2 + 3 + 4 = 10 $
# Answer:
$ 10 $

### Problem 42: $\sum_{k=1}^{4} k^2$
# Explanation:
## Step1: Expand the summation
Substitute $ k = 1, 2, 3, 4 $ into the term $ k^2 $:
$ \sum_{k=1}^{4} k^2 = 1^2 + 2^2 + 3^2 + 4^2 $
## Step2: Calculate each square
$ 1^2 = 1 $, $ 2^2 = 4 $, $ 3^2 = 9 $, $ 4^2 = 16 $
## Step3: Sum the results
$ 1 + 4 + 9 + 16 = 30 $
# Answer:
$ 30 $

### Problem 43: $\sum_{k=1}^{3} \frac{1}{k}$
# Explanation:
## Step1: Expand the summation
Substitute $ k = 1, 2, 3 $ into the term $ \frac{1}{k} $:
$ \sum_{k=1}^{3} \frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} $
## Step2: Find a common denominator and add
The common denominator of 1, 2, 3 is 6:
$ \frac{6}{6} + \frac{3}{6} + \frac{2}{6} = \frac{6 + 3 + 2}{6} = \frac{11}{6} $
# Answer:
$ \frac{11}{6} $

### Problem 44: $\sum_{j=1}^{100} (-1)^j$
# Explanation:
## Step1: Analyze the pattern
For $ j = 1 $, $ (-1)^1 = -1 $; $ j = 2 $, $ (-1)^2 = 1 $; $ j = 3 $, $ (-1)^3 = -1 $; $ j = 4 $, $ (-1)^4 = 1 $; and so on.
## Step2: Group the terms
Pair the terms: $ (-1 + 1) + (-1 + 1) + \dots + (-1 + 1) $ (50 pairs, since 100 terms)
## Step3: Calculate each pair
Each pair $ (-1 + 1) = 0 $, so 50 pairs sum to $ 50 \times 0 = 0 $
# Answer:
$ 0 $

### Problem 45: $\sum_{i=1}^{8} [1 + (-1)^i]$
# Explanation:
## Step1: Analyze the term for even and odd $ i $
- If $ i $ is odd ($ i = 1, 3, 5, 7 $): $ (-1)^i = -1 $, so $ 1 + (-1) = 0 $
- If $ i $ is even ($ i = 2, 4, 6, 8 $): $ (-1)^i = 1 $, so $ 1 + 1 = 2 $
## Step2: Count the number of odd and even terms
There are 4 odd $ i $ and 4 even $ i $ in $ i = 1 $ to $ 8 $
## Step3: Calculate the sum
Sum of odd terms: $ 4 \times 0 = 0 $; Sum of even terms: $ 4 \times 2 = 8 $; Total sum: $ 0 + 8 = 8 $
# Answer:
$ 8 $

### Problem 46: $\sum_{i=4}^{12} 10$
# Explanation:
## Step1: Determine the number of terms
The number of terms from $ i = 4 $ to $ i = 12 $ is $ 12 - 4 + 1 = 9 $
## Step2: Calculate the sum
Each term is 10, so the sum is $ 9 \times 10 = 90 $
# Answer:
$ 90 $

### Problem 47: $\sum_{k=1}^{5} 2^{k - 1}$
# Explanation:
## Step1: Expand the summation
Substitute $ k = 1, 2, 3, 4, 5 $ into $ 2^{k - 1} $:
$ 2^{0} + 2^{1} + 2^{2} + 2^{3} + 2^{4} $
## Step2: Calculate each term
$ 2^0 = 1 $, $ 2^1 = 2 $, $ 2^2 = 4 $, $ 2^3 = 8 $, $ 2^4 = 16 $
## Step3: Sum the terms
$ 1 + 2 + 4 + 8 + 16 = 31 $
# Answer:
$ 31 $

### Problem 48: $\sum_{i=1}^{3} i2^i$
# Explanation:
## Step1: Expand the summation
Substitute $ i = 1, 2, 3 $ into $ i2^i $:
$ 1 \times 2^1 + 2 \times 2^2 + 3 \times 2^3 $
## Step2: Calculate each term
$ 1 \times 2 = 2 $, $ 2 \times 4 = 8 $, $ 3 \times 8 = 24 $
## Step3: Sum the terms
$ 2 + 8 + 24 = 34 $
# Answer:
$ 34 $

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QUESTION IMAGE

Question

Explanation:

Problem 41: $\sum_{k=1}^{4} k$

Step1: Expand the summation

Step2: Calculate the sum

Step1: Expand the summation

Step2: Calculate each square

Step3: Sum the results

Step1: Expand the summation

Step2: Find a common denominator and add

Answer:

Problem 42: $\sum_{k=1}^{4} k^2$