Evaluate the summation below. 5sum_(k=0)^(3)(1-3k^(2)) Answer:

Write Terms Explicitly: Write out the terms of the series explicitly. The series is 5 times the sum from k=0 to 3 of (1-3k^2). This means we need to calculate the value of (1-3k^2) for each k from 0 to 3 and then multiply the sum of these values by 5.Calculate Terms: Calculate the terms of the series. For k=0: (1-3(0)^2) = 1 For k=1: (1-3(1)^2) = 1-3(1) = 1-3 = -2 For k=2: (1-3(2)^2) = 1-3(4) = 1-12 = -11 For k=3: (1-3(3)^2) = 1-3(9) = 1-27 = -26Sum Terms: Sum the terms of the series. The sum of the terms (1-3k^2) from k=0 to 3 is 1 + (-2) + (-11) + (-26). This sum is 1 - 2 - 11 - 26 = -38.Multiply by 5: Multiply the sum by 5. The final step is to multiply the sum of the series by 5. 5 * (-38) = -190.

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Evaluate the summation below.

5sum_(k=0)^(3)(1-3k^(2))
Answer:

Evaluate the summation below. $\newline$ $5 \sum_{k=0}^{3}\left(1-3 k^{2}\right)$ $\newline$ Answer:

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Algebra 2

Sum of finite series starts from 1

Full solution

Q. Evaluate the summation below.

\newline

5 \sum_{k=0}^{3}\left(1-3 k^{2}\right)

\newline

Answer:

Write Terms Explicitly: Write out the terms of the series explicitly. $\newline$ The series is $5$ times the sum from $k=0$ to $3$ of $(1-3k^2)$ . This means we need to calculate the value of $(1-3k^2)$ for each $k$ from $0$ to $3$ and then multiply the sum of these values by $5$ .
Calculate Terms: Calculate the terms of the series. $\newline$ For $k=0$ : $(1-3(0)^2) = 1$ $\newline$ For $k=1$ : $(1-3(1)^2) = 1-3(1) = 1-3 = -2$ $\newline$ For $k=2$ : $(1-3(2)^2) = 1-3(4) = 1-12 = -11$ $\newline$ For $k=3$ : $(1-3(3)^2) = 1-3(9) = 1-27 = -26$
Sum Terms: Sum the terms of the series. $\newline$ The sum of the terms $1-3k^2$ from $k=0$ to $3$ is $1 + (-2) + (-11) + (-26)$ . $\newline$ This sum is $1 - 2 - 11 - 26 = -38$ .
Multiply by $5$ : Multiply the sum by $5$ . The final step is to multiply the sum of the series by $5$ . $5 \times (-38) = -190$ .