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Find the sum.

sum_(k=1)^(38)(6k-105)=

Find the sum. $\newline$ $\sum_{k=1}^{38}(6 k-105)=$

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Find derivatives using the chain rule I

Full solution

Q. Find the sum.

\newline

\sum_{k=1}^{38}(6 k-105)=

Find Sum of $6k$ : We need to find the sum of the series given by the expression $(6k-105)$ from $k=1$ to $38$ . We can separate the series into two parts: the sum of $6k$ and the sum of $-105$ .
Calculate Sum of $-105$ : First, let's find the sum of the series $6k$ from $k=1$ to $38$ . This is an arithmetic series with a common difference of $6$ . 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.
Find Total Sum: The first term $a_1$ when $k=1$ is $6\times1 = 6$ , and the last term $a_n$ when $k=38$ is $6\times38 = 228$ . There are $38$ terms in total. Plugging these values into the formula gives us $S = \frac{38}{2} \times (6 + 228)$ .
Find Total Sum: The first term $a_1$ when $k=1$ is $6\times1 = 6$ , and the last term $a_n$ when $k=38$ is $6\times38 = 228$ . There are $38$ terms in total. Plugging these values into the formula gives us $S = \frac{38}{2} \times (6 + 228)$ .Calculating the sum $S = 19 \times (6 + 228) = 19 \times 234 = 4446$ . This is the sum of the $6k$ part of the series.
Find Total Sum: The first term $a_1$ when $k=1$ is $6\times1 = 6$ , and the last term $a_n$ when $k=38$ is $6\times38 = 228$ . There are $38$ terms in total. Plugging these values into the formula gives us $S = \frac{38}{2} \times (6 + 228)$ .Calculating the sum $S = 19 \times (6 + 228) = 19 \times 234 = 4446$ . This is the sum of the $6k$ part of the series.Now, let's find the sum of the constant part of the series, which is $k=1$ $0$ repeated $38$ times. The sum of a constant series is simply the constant times the number of terms. So, the sum is $k=1$ $2$ .
Find Total Sum: The first term $a_1$ when $k=1$ is $6\times1 = 6$ , and the last term $a_n$ when $k=38$ is $6\times38 = 228$ . There are $38$ terms in total. Plugging these values into the formula gives us $S = \frac{38}{2} \times (6 + 228)$ .Calculating the sum $S = 19 \times (6 + 228) = 19 \times 234 = 4446$ . This is the sum of the $6k$ part of the series.Now, let's find the sum of the constant part of the series, which is $k=1$ $0$ repeated $38$ times. The sum of a constant series is simply the constant times the number of terms. So, the sum is $k=1$ $2$ .Calculating the sum of the constant part gives us $k=1$ $3$ .
Find Total Sum: The first term $a_1$ when $k=1$ is $6\times1 = 6$ , and the last term $a_n$ when $k=38$ is $6\times38 = 228$ . There are $38$ terms in total. Plugging these values into the formula gives us $S = \frac{38}{2} \times (6 + 228)$ . Calculating the sum $S = 19 \times (6 + 228) = 19 \times 234 = 4446$ . This is the sum of the $6k$ part of the series. Now, let's find the sum of the constant part of the series, which is $k=1$ $0$ repeated $38$ times. The sum of a constant series is simply the constant times the number of terms. So, the sum is $k=1$ $2$ . Calculating the sum of the constant part gives us $k=1$ $3$ . Finally, we add the sum of the $6k$ part and the sum of the constant part to get the total sum of the series. So, the total sum is $k=1$ $5$ .

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