Find the numerical answer to the summation given below. sum_(n=0)^(89)(3n+8) Answer:

Calculate number of terms: We need to find the sum of the arithmetic series given by the formula (3n+8) where n ranges from 0 to 89. The general formula for the sum of an arithmetic series is S = (n/2)(a_1 + a_n), where n is the number of terms, a_1 is the first term, and a_n is the last term.Find first term: First, we calculate the number of terms in the series. Since the series starts at n=0 and ends at n=89, there are 89 - 0 + 1 = 90 terms.Find last term: Next, we find the first term of the series by substituting n=0 into the formula (3n+8), which gives us a_1 = 3(0) + 8 = 8.Use sum formula: Then, we find the last term of the series by substituting n=89 into the formula (3n+8), which gives us a_n = 3(89) + 8 = 267 + 8 = 275.Perform calculation: Now we can use the sum formula for an arithmetic series: S = (n/2)(a_1 + a_n). Substituting the values we have, S = (90/2)(8 + 275).Multiply terms: Perform the calculation inside the parentheses first: 8 + 275 = 283.Find sum: Now multiply the number of terms divided by 2 with the sum of the first and last term: S = (90/2) * 283 = 45 * 283.Finally, perform the multiplication to find the sum of the series: S = 45 * 283 = 12735.

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Algebra 2

Sum of finite series starts from 1

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Q. Find the numerical answer to the summation given below.

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\sum_{n=0}^{89}(3 n+8)

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Answer:

Calculate number of terms: We need to find the sum of the arithmetic series given by the formula $(3n+8)$ where $n$ ranges from $0$ to $89$ . The general formula for the sum of an arithmetic series is $S = \frac{n}{2}(a_1 + a_n)$ , where $n$ is the number of terms, $a_1$ is the first term, and $a_n$ is the last term.
Find first term: First, we calculate the number of terms in the series. Since the series starts at $n=0$ and ends at $n=89$ , there are $89 - 0 + 1 = 90$ terms.
Find last term: Next, we find the first term of the series by substituting $n=0$ into the formula $(3n+8)$ , which gives us $a_1 = 3(0) + 8 = 8$ .
Use sum formula: Then, we find the last term of the series by substituting $n=89$ into the formula $(3n+8)$ , which gives us $a_n = 3(89) + 8 = 267 + 8 = 275$ .
Perform calculation: Now we can use the sum formula for an arithmetic series: $S = \frac{n}{2}(a_1 + a_n)$ . Substituting the values we have, $S = \frac{90}{2}(8 + 275)$ .
Multiply terms: Perform the calculation inside the parentheses first: $8 + 275 = 283$ .
Find sum: Now multiply the number of terms divided by $2$ with the sum of the first and last term: $S = (\frac{90}{2}) \times 283 = 45 \times 283$ .
Find sum: Now multiply the number of terms divided by $2$ with the sum of the first and last term: $S = \frac{90}{2} \times 283 = 45 \times 283$ .Finally, perform the multiplication to find the sum of the series: $S = 45 \times 283 = 12735$ .