Question

find the sum by adding each term together. use the summation capabilities of a graphing utility to verify your result. (round your answer to four decimal places.)
\\(\sum_{k=3}^{7} \frac{1}{k^2 + 3}\\)
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Explanation:

Step1: Identify terms of the sum

The sum is $\sum_{k=3}^{7} \frac{1}{k^2 + 3}$, so $k$ takes values 3,4,5,6,7.

Step2: Calculate each term

• For $k=3$: $\frac{1}{3^2 + 3} = \frac{1}{9+3} = \frac{1}{12} \approx 0.083333$
• For $k=4$: $\frac{1}{4^2 + 3} = \frac{1}{16+3} = \frac{1}{19} \approx 0.052632$
• For $k=5$: $\frac{1}{5^2 + 3} = \frac{1}{25+3} = \frac{1}{28} \approx 0.035714$
• For $k=6$: $\frac{1}{6^2 + 3} = \frac{1}{36+3} = \frac{1}{39} \approx 0.025641$
• For $k=7$: $\frac{1}{7^2 + 3} = \frac{1}{49+3} = \frac{1}{52} \approx 0.019231$

Step3: Sum all calculated terms

$0.083333 + 0.052632 + 0.035714 + 0.025641 + 0.019231$

Step4: Round to 4 decimal places

Add the values: $0.083333+0.052632=0.135965$; $0.135965+0.035714=0.171679$; $0.171679+0.025641=0.19732$; $0.19732+0.019231=0.216551$. Round to 4 decimals.

Answer:

0.2166