Find the numerical answer to the summation given below. sum_(n=0)^(64)(2n+5) Answer:

Calculate number of terms: We need to find the sum of the arithmetic series where each term is given by the formula (2n+5), starting from n=0 and ending at n=64. 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 we start at n=0 and go to n=64, we have 64 - 0 + 1 = 65 terms.Find last term: Next, we find the first term of the series by substituting n=0 into the formula (2n+5), which gives us a_1 = 2(0) + 5 = 5.Use sum formula: Then, we find the last term of the series by substituting n=64 into the formula (2n+5), which gives us a_64 = 2(64) + 5 = 128 + 5 = 133.Calculate sum: Now we can use the sum formula for an arithmetic series: S = (n/2)(a_1 + a_n). Substituting the values we have, S = (65/2)(5 + 133).Multiply terms: We calculate the sum inside the parentheses: 5 + 133 = 138.Final calculation: Finally, we multiply the number of terms by the average of the first and last term: S = (65/2) * 138.Performing the multiplication, we get S = 65 * 69 = 4485.

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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}^{64}(2 n+5)

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Answer:

Calculate number of terms: We need to find the sum of the arithmetic series where each term is given by the formula $(2n+5)$ , starting from $n=0$ and ending at $n=64$ . 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 we start at $n=0$ and go to $n=64$ , we have $64 - 0 + 1 = 65$ terms.
Find last term: Next, we find the first term of the series by substituting $n=0$ into the formula $(2n+5)$ , which gives us $a_1 = 2(0) + 5 = 5$ .
Use sum formula: Then, we find the last term of the series by substituting $n=64$ into the formula $(2n+5)$ , which gives us $a_{64} = 2(64) + 5 = 128 + 5 = 133$ .
Calculate sum: Now we can use the sum formula for an arithmetic series: $S = \frac{n}{2}(a_1 + a_n)$ . Substituting the values we have, $S = \frac{65}{2}(5 + 133)$ .
Multiply terms: We calculate the sum inside the parentheses: $5 + 133 = 138$ .
Final calculation: Finally, we multiply the number of terms by the average of the first and last term: $S = \left(\frac{65}{2}\right) \times 138$ .
Final calculation: Finally, we multiply the number of terms by the average of the first and last term: $S = \frac{65}{2} \times 138$ . Performing the multiplication, we get $S = 65 \times 69 = 4485$ .