Find the numerical answer to the summation given below. sum_(n=3)^(79)(6n+4) Answer:

Split Summation: Recognize that the summation of the series can be split into two separate summations, one for 6n and one for 4. This gives us sum_(n=3)^(79)6n + sum_(n=3)^(79)4.Evaluate First Term: Evaluate the first term sum_(n=3)^(79)6n. This is an arithmetic series, and the sum of an arithmetic series can be found using the formula 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. The first term when n=3 is 6*3 = 18, and the last term when n=79 is 6*79 = 474. The number of terms is 79 - 3 + 1 = 77.Calculate First Term Sum: Calculate the sum of the first term sum_(n=3)^(79)6n using the arithmetic series sum formula. S = 77/2 * (18 + 474) = 38.5 * 492 = 18954Evaluate Second Term: Evaluate the second term sum_(n=3)^(79)4. This is a constant series, and the sum of a constant series is simply the constant times the number of terms. We already found the number of terms to be 77 in Step 2.Calculate Second Term Sum: Calculate the sum of the second term sum_(n=3)^(79)4. S = 4 * 77 = 308Add Results: Add the results from Step 3 and Step 5 to get the final answer. Total Sum = 18954 + 308 = 19262

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Find the numerical answer to the summation given below.

sum_(n=3)^(79)(6n+4)
Answer:

Find the numerical answer to the summation given below. $\newline$ $\sum_{n=3}^{79}(6 n+4)$ $\newline$ Answer:

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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.

\newline

\sum_{n=3}^{79}(6 n+4)

\newline

Answer:

Split Summation: Recognize that the summation of the series can be split into two separate summations, one for $6n$ and one for $4$ . This gives us $\sum_{n=3}^{79}6n + \sum_{n=3}^{79}4$ .
Evaluate First Term: Evaluate the first term $\sum_{n=3}^{79}6n$ . This is an arithmetic series, and the sum of an arithmetic series can be found using the formula $S = \frac{n}{2} \times (a_1 + a_n)$ , where $n$ is the number of terms, $a_1$ is the first term, and $a_n$ is the last term. The first term when $n=3$ is $6\times3 = 18$ , and the last term when $n=79$ is $6\times79 = 474$ . The number of terms is $79 - 3 + 1 = 77$ .
Calculate First Term Sum: Calculate the sum of the first term $\sum_{n=3}^{79} 6n$ using the arithmetic series sum formula. $\newline$ $S = \frac{77}{2} \times (18 + 474)$ $\newline$ $= 38.5 \times 492$ $\newline$ $= 18954$
Evaluate Second Term: Evaluate the second term $\sum_{n=3}^{79} 4$ . This is a constant series, and the sum of a constant series is simply the constant times the number of terms. We already found the number of terms to be $77$ in Step $2$ .
Calculate Second Term Sum: Calculate the sum of the second term $\sum_{n=3}^{79}4$ . $S = 4 \times 77$ $= 308$
Add Results: Add the results from Step $3$ and Step $5$ to get the final answer. $\newline$ Total Sum = $18954 + 308$ $\newline$ = $19262$