Find the numerical answer to the summation given below. sum_(n=0)^(72)(6n+3) Answer:

Calculate number of terms: We need to find the sum of the arithmetic series given by the formula (6n+3) from n=0 to n=72. 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=72, we have 72 - 0 + 1 = 73 terms.Find last term: Next, we find the first term of the series by substituting n=0 into the formula (6n+3). This gives us a_1 = 6*0 + 3 = 3.Use sum formula: Now, we find the last term of the series by substituting n=72 into the formula (6n+3). This gives us a_72 = 6*72 + 3 = 435.Perform calculation: We can now use the sum formula for an arithmetic series: S = n/2 * (a_1 + a_n). Substituting the values we have, S = 73/2 * (3 + 435).Multiply by 73/2: Perform the calculation inside the parentheses first: 3 + 435 = 438.Perform final multiplication: Now, multiply 438 by 73/2 to find the sum of the series: S = 73/2 * 438 = 36.5 * 438.Finally, perform the multiplication to find the sum: S = 36.5 * 438 = 15987.

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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}^{72}(6 n+3)

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Answer:

Calculate number of terms: We need to find the sum of the arithmetic series given by the formula $6n+3$ from $n=0$ to $n=72$ . The general formula for the sum of an arithmetic series is $S = \frac{n}{2} \times (a_1 + a_n)$ , 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 $n=72$ , we have $72 - 0 + 1 = 73$ terms.
Find last term: Next, we find the first term of the series by substituting $n=0$ into the formula $(6n+3)$ . This gives us $a_1 = 6\times 0 + 3 = 3$ .
Use sum formula: Now, we find the last term of the series by substituting $n=72$ into the formula $(6n+3)$ . This gives us $a_{72} = 6\times72 + 3 = 435$ .
Perform calculation: We can now use the sum formula for an arithmetic series: $S = \frac{n}{2} \times (a_1 + a_n)$ . Substituting the values we have, $S = \frac{73}{2} \times (3 + 435)$ .
Multiply by $\frac{73}{2}$ : Perform the calculation inside the parentheses first: $3 + 435 = 438$ .
Perform final multiplication: Now, multiply $438$ by $\frac{73}{2}$ to find the sum of the series: $S = \frac{73}{2} \times 438 = 36.5 \times 438$ .
Perform final multiplication: Now, multiply $438$ by $\frac{73}{2}$ to find the sum of the series: $S = \frac{73}{2} \times 438 = 36.5 \times 438$ .Finally, perform the multiplication to find the sum: $S = 36.5 \times 438 = 15987$ .