Find the sum. sum_(k=1)^(40)(5k-87)=

Find Sum of 5k: We need to find the sum of the series given by the expression (5k-87) from k=1 to k=40. We can separate the series into two parts: the sum of 5k and the sum of -87.Find Sum of -87: First, let's find the sum of the series 5k from k=1 to k=40. This is an arithmetic series with a common difference of 5. The sum of an arithmetic series can be found using the formula S = n/2 * (a1 + an), where n is the number of terms, a1 is the first term, and an is the last term.Calculate Total Sum: The first term a1 when k=1 is 5(1) = 5, and the last term an when k=40 is 5(40) = 200. The number of terms n is 40. Plugging these values into the formula gives us S = 40/2 * (5 + 200).Calculating the sum S, we get S = 20 * (205) = 4100. This is the sum of the series 5k from k=1 to k=40.Now, let's find the sum of the series -87 from k=1 to k=40. Since -87 is a constant, the sum of this series is simply -87 multiplied by the number of terms, which is 40.Calculating the sum of -87 for 40 terms, we get -87 * 40 = -3480. This is the sum of the series -87 from k=1 to k=40.To find the total sum of the original series (5k-87) from k=1 to k=40, we add the sum of the series 5k and the sum of the series -87. This gives us 4100 + (-3480).Calculating the total sum, we get 4100 - 3480 = 620. This is the final answer for the sum of the series (5k-87) from k=1 to k=40.

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Find the sum. $\newline$ $\sum_{k=1}^{40}(5 k-87)=$

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Grade 8

Powers with negative bases

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Q. Find the sum.

\newline

\sum_{k=1}^{40}(5 k-87)=

Find Sum of $5k$ : We need to find the sum of the series given by the expression $(5k-87)$ from $k=1$ to $k=40$ . We can separate the series into two parts: the sum of $5k$ and the sum of $-87$ .
Find Sum of $-87$ : First, let's find the sum of the series $5k$ from $k=1$ to $k=40$ . This is an arithmetic series with a common difference of $5$ . 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.
Calculate Total Sum: The first term $a_1$ when $k=1$ is $5(1) = 5$ , and the last term $a_n$ when $k=40$ is $5(40) = 200$ . The number of terms $n$ is $40$ . Plugging these values into the formula gives us $S = \frac{40}{2} \times (5 + 200)$ .
Calculate Total Sum: The first term $a_1$ when $k=1$ is $5(1) = 5$ , and the last term $a_n$ when $k=40$ is $5(40) = 200$ . The number of terms $n$ is $40$ . Plugging these values into the formula gives us $S = \frac{40}{2} \cdot (5 + 200)$ . Calculating the sum $S$ , we get $k=1$ $0$ . This is the sum of the series $k=1$ $1$ from $k=1$ to $k=40$ .
Calculate Total Sum: The first term $a_1$ when $k=1$ is $5(1) = 5$ , and the last term $a_n$ when $k=40$ is $5(40) = 200$ . The number of terms $n$ is $40$ . Plugging these values into the formula gives us $S = \frac{40}{2} \times (5 + 200)$ . Calculating the sum $S$ , we get $k=1$ $0$ . This is the sum of the series $k=1$ $1$ from $k=1$ to $k=40$ . Now, let's find the sum of the series $k=1$ $4$ from $k=1$ to $k=40$ . Since $k=1$ $4$ is a constant, the sum of this series is simply $k=1$ $4$ multiplied by the number of terms, which is $40$ .
Calculate Total Sum: The first term $a_1$ when $k=1$ is $5(1) = 5$ , and the last term $a_n$ when $k=40$ is $5(40) = 200$ . The number of terms $n$ is $40$ . Plugging these values into the formula gives us $S = \frac{40}{2} \times (5 + 200)$ . Calculating the sum $S$ , we get $k=1$ $0$ . This is the sum of the series $k=1$ $1$ from $k=1$ to $k=40$ . Now, let's find the sum of the series $k=1$ $4$ from $k=1$ to $k=40$ . Since $k=1$ $4$ is a constant, the sum of this series is simply $k=1$ $4$ multiplied by the number of terms, which is $40$ . Calculating the sum of $k=1$ $4$ for $40$ terms, we get $5(1) = 5$ $2$ . This is the sum of the series $k=1$ $4$ from $k=1$ to $k=40$ .
Calculate Total Sum: The first term $a_1$ when $k=1$ is $5(1) = 5$ , and the last term $a_n$ when $k=40$ is $5(40) = 200$ . The number of terms $n$ is $40$ . Plugging these values into the formula gives us $S = \frac{40}{2} \times (5 + 200)$ . Calculating the sum $S$ , we get $k=1$ $0$ . This is the sum of the series $k=1$ $1$ from $k=1$ to $k=40$ . Now, let's find the sum of the series $k=1$ $4$ from $k=1$ to $k=40$ . Since $k=1$ $4$ is a constant, the sum of this series is simply $k=1$ $4$ multiplied by the number of terms, which is $40$ . Calculating the sum of $k=1$ $4$ for $40$ terms, we get $5(1) = 5$ $2$ . This is the sum of the series $k=1$ $4$ from $k=1$ to $k=40$ . To find the total sum of the original series $5(1) = 5$ $6$ from $k=1$ to $k=40$ , we add the sum of the series $k=1$ $1$ and the sum of the series $k=1$ $4$ . This gives us $a_n$ $1$ .
Calculate Total Sum: The first term $a_1$ when $k=1$ is $5(1) = 5$ , and the last term $a_n$ when $k=40$ is $5(40) = 200$ . The number of terms $n$ is $40$ . Plugging these values into the formula gives us $S = \frac{40}{2} \times (5 + 200)$ . Calculating the sum $S$ , we get $k=1$ $0$ . This is the sum of the series $k=1$ $1$ from $k=1$ to $k=40$ . Now, let's find the sum of the series $k=1$ $4$ from $k=1$ to $k=40$ . Since $k=1$ $4$ is a constant, the sum of this series is simply $k=1$ $4$ multiplied by the number of terms, which is $40$ . Calculating the sum of $k=1$ $4$ for $40$ terms, we get $5(1) = 5$ $2$ . This is the sum of the series $k=1$ $4$ from $k=1$ to $k=40$ . To find the total sum of the original series $5(1) = 5$ $6$ from $k=1$ to $k=40$ , we add the sum of the series $k=1$ $1$ and the sum of the series $k=1$ $4$ . This gives us $a_n$ $1$ . Calculating the total sum, we get $a_n$ $2$ . This is the final answer for the sum of the series $5(1) = 5$ $6$ from $k=1$ to $k=40$ .

Find the sum. $\newline$ $\sum_{k=1}^{40}(5 k-87)=$

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Find the sum.\newline∑k=140(5k−87)= \sum_{k=1}^{40}(5 k-87)= k=1∑40​(5k−87)=

Find the sum. $\newline$ $\sum_{k=1}^{40}(5 k-87)=$