Evaluate the summation below. 6sum_(p=0)^(3)(-p^(2)+2p) Answer:

Expand and Calculate: We start by expanding the summation expression -p^2 + 2p for each value of p from 0 to 3. For p = 0: -0^2 + 2*0 = 0 For p = 1: -1^2 + 2*1 = -1 + 2 = 1 For p = 2: -2^2 + 2*2 = -4 + 4 = 0 For p = 3: -3^2 + 2*3 = -9 + 6 = -3Sum up Results: Now, we sum up the results of the expression for each value of p. Sum: 0 + 1 + 0 - 3 = -2Multiply by 6: Finally, we multiply the sum by 6 as indicated by the summation expression 6sum_(p=0)^(3)(-p^2 + 2p). 6 * (-2) = -12

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Evaluate the summation below.

6sum_(p=0)^(3)(-p^(2)+2p)
Answer:

Evaluate the summation below. $\newline$ $6 \sum_{p=0}^{3}\left(-p^{2}+2 p\right)$ $\newline$ Answer:

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Algebra 2

Sum of finite series starts from 1

Full solution

Q. Evaluate the summation below.

\newline

6 \sum_{p=0}^{3}\left(-p^{2}+2 p\right)

\newline

Answer:

Expand and Calculate: We start by expanding the summation expression $-p^2 + 2p$ for each value of $p$ from $0$ to $3$ . $\newline$ For $p = 0$ : $-0^2 + 2\cdot0 = 0$ $\newline$ For $p = 1$ : $-1^2 + 2\cdot1 = -1 + 2 = 1$ $\newline$ For $p = 2$ : $-2^2 + 2\cdot2 = -4 + 4 = 0$ $\newline$ For $p$ $0$ : $p$ $1$
Sum up Results: Now, we sum up the results of the expression for each value of $p$ . $\newline$ Sum: $0 + 1 + 0 - 3 = -2$
Multiply by $6$ : Finally, we multiply the sum by $6$ as indicated by the summation expression $6\sum_{p=0}^{3}(-p^2 + 2p)$ . $\newline$ $6 \times (-2) = -12$