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Nady [450]
3 years ago
9

Write out the following sums, one term for each value of k. Simplify each term as much as possible, but do not enter decimals. F

or example, enter 1+4+9 instead of 12+22+32 or 14, or enter 1/2+1/2 instead of 0.5+0.5 or 1. The purpose of this problem is for you to show that you know how to interpret summation notation and write all of the terms in a sum, which is why you are being told not to reduce your answers very much.
Mathematics
1 answer:
alukav5142 [94]3 years ago
7 0

The correct question is:

Write out the following sums, one term for each value of k. Simplify each term as much as possible, but do not enter decimals. For example, enter 1 + 4 + 9 instead of 1² + 2² + 3² or 14, or enter 1/2 + 1/2 instead of 0.5 + 0.5 or 1.

The purpose of this problem is for you to show that you know how to interpret summation notation and write all of the terms in a sum, which is why you are being told not to reduce your answers very much.

(a)  \sum_{k=0}^5 2^k \\ \\(b)   \sum_{k=2}^7 \frac{1}{k} \\ \\(c)  \sum_{k=1}^5 k^2 \\ \\(d)  \sum_{k=1}^6 \frac{1}{6} \\ \\(e)  \sum_{k=1}^6 2k

Answer:

(a)  \sum_{k=0}^5 2^k = $1 + 2 + 4 + 8 + 16 + 32$ \\ \\(b)   \sum_{k=2}^7 \frac{1}{k} = \frac{1}{2}  + \frac{1}{3} + \frac{1}{4}+ \frac{1}{5}+ \frac{1}{6}+ \frac{1}{7}  \\ \\(c)  \sum_{k=1}^5 k^2 = 1 + 4 + 9 + 16 + 25 \\ \\(d)  \sum_{k=1}^6 \frac{1}{6} = \frac{1}{6}  + \frac{1}{6}  + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} \\ \\(e)  \sum_{k=1}^6 2k = 2 +4 +6 +8 +10 +12

Step-by-step explanation:

(a) \sum_{k=0}^5 2^k\\For k = 0: 2^k = 2^0 = 1\\For k = 1: 2^1 = 2\\For k = 2: 2^2 = 4\\For k = 3: 2^3 = 8\\For k = 4: 2^4 = 16\\For k = 5: 2^5 = 32\\\sum_{k=0}^5 2^k = 1 + 2 + 4 + 8 + 16 + 32

(b) \sum_{k=2}^7 \frac{1}{k}\\For k = 2: 1/2\\For k = 3: 1/3\\For k = 4: 1/4\\For k = 5: 1/5\\For k = 6: 1/6\\For k = 7: 1/7\\ \sum_{k=2}^7 \frac{1}{k} = 1/2 + 1/3 + 1/4 + 1/5 + 1/6+ 1/7

(c)  \sum_{k=1}^5 k^2\\For k = 1: 1^2 = 1\\For k = 2: 2^2 = 4\\For k = 3: 3^2 = 9\\For k = 4: 4^2= 16\\For k = 5: 5^2 = 25\\\sum_{k=1}^5 k^2 = 1 + 4 + 9 + 16 + 25

(d) \sum_{k=1}^6 \frac{1}{6}\\For k = 1: 1/6\\For k = 2: 1/6\\For k = 3: 1/6\\For k = 4: 1/6\\For k = 5: 1/6\\For k = 6: 1/6\\ \sum_{k=1}^6 \frac{1}{6} = 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6

(e) \sum_{k=1}^6 2k\\For k = 1: 2\times1 = 2\\For k = 2: 2\times2 = 4\\For k = 3: 2\times3 = 6\\For k = 4: 2\times4 = 8\\For k = 5: 2\times5 = 10\\For k = 6: 2\times6 = 12\\\sum_{k=1}^6 2k = 2 +4 +6 +8 +10 +12

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