Find the numerical answer to the summation given below. sum_(n=1)^(70)(3n+10) Answer:

Recognize Split Summations: Recognize that the summation of the series can be split into two separate summations: the summation of 3n from n=1 to 70 and the summation of 10 from n=1 to 70. This can be written as sum_(n=1)^(70)(3n) + sum_(n=1)^(70)(10).Calculate 3n Summation: Calculate the first summation sum_(n=1)^(70)(3n). This is an arithmetic series where each term is 3 times the corresponding term of the sum of the first 70 natural numbers. The sum of the first m natural numbers is given by the formula m(m+1)/2. Therefore, the sum of 3n from 1 to 70 is 3 * (70(70+1)/2).Calculate 3n Sum: Perform the calculation from Step 2. 3 * (70(70+1)/2) = 3 * (70*71/2) = 3 * (4970) = 14910.Calculate 10 Summation: Calculate the second summation sum_(n=1)^(70)(10). Since 10 is a constant, the sum is simply 10 added to itself 70 times, which is 10*70.Calculate 10 Sum: Perform the calculation from Step 4. 10*70 = 700.Find Total Sum: Add the results from Step 3 and Step 5 to find the total sum of the series. 14910 + 700 = 15610.

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Find the numerical answer to the summation given below.

sum_(n=1)^(70)(3n+10)
Answer:

Find the numerical answer to the summation given below. $\newline$ $\sum_{n=1}^{70}(3 n+10)$ $\newline$ 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=1}^{70}(3 n+10)

\newline

Answer:

Recognize Split Summations: Recognize that the summation of the series can be split into two separate summations: the summation of $3n$ from $n=1$ to $70$ and the summation of $10$ from $n=1$ to $70$ . This can be written as $\sum_{n=1}^{70}(3n) + \sum_{n=1}^{70}(10)$ .
Calculate $3n$ Summation: Calculate the first summation $\sum_{n=1}^{70}(3n)$ . This is an arithmetic series where each term is $3$ times the corresponding term of the sum of the first $70$ natural numbers. The sum of the first $m$ natural numbers is given by the formula $\frac{m(m+1)}{2}$ . Therefore, the sum of $3n$ from $1$ to $70$ is $3 \times \left(\frac{70(70+1)}{2}\right)$ .
Calculate $3n$ Sum: Perform the calculation from Step $2$ . $\newline$ $3 \times \left(\frac{70(70+1)}{2}\right) = 3 \times \left(\frac{70\times71}{2}\right) = 3 \times (4970) = 14910$ .
Calculate $10$ Summation: Calculate the second summation $\sum_{n=1}^{70}(10)$ . Since $10$ is a constant, the sum is simply $10$ added to itself $70$ times, which is $10\times70$ .
Calculate $10$ Sum: Perform the calculation from Step $4$ . $\newline$ $10\times70 = 700$ .
Find Total Sum: Add the results from Step $3$ and Step $5$ to find the total sum of the series. $14910 + 700 = 15610$ .