Find the numerical answer to the summation given below. sum_(n=0)^(68)(6n+3) Answer:

Separate Summation Parts: To find the sum of the series, we can separate the summation into two parts: the summation of 6n from n=0 to 68 and the summation of 3 from n=0 to 68.Find Sum of 6n Series: First, let's find the sum of the arithmetic series 6n. The sum of an arithmetic series is given by the formula S = n/2 * (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 6n Part: The first term when n=0 is 6*0 = 0, and the last term when n=68 is 6*68 = 408. There are 69 terms in total because we start counting from n=0.Calculate Constant Term: Now we can calculate the sum of the 6n part: S_6n = 69/2 * (0 + 408) = 69/2 * 408 = 34.5 * 408 = 14076.Calculate Total Sum: Next, we calculate the sum of the constant term 3, which is added 69 times (once for each value of n from 0 to 68). This sum is simply 3 * 69.Calculating the sum of the constant term: S_3 = 3 * 69 = 207.Finally, we add the two sums together to find the total sum of the series: Total Sum = S_6n + S_3 = 14076 + 207.Adding the two sums gives us the final answer: Total Sum = 14283.

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Find the numerical answer to the summation given below.

sum_(n=0)^(68)(6n+3)
Answer:

Find the numerical answer to the summation given below. $\newline$ $\sum_{n=0}^{68}(6 n+3)$ $\newline$ Answer:

View step-by-step help

Home

Math Problems

Precalculus

Find the roots of factored polynomials

Full solution

Q. Find the numerical answer to the summation given below.

\newline

\sum_{n=0}^{68}(6 n+3)

\newline

Answer:

Separate Summation Parts: To find the sum of the series, we can separate the summation into two parts: the summation of $6n$ from $n=0$ to $68$ and the summation of $3$ from $n=0$ to $68$ .
Find Sum of $6$ n Series: First, let's find the sum of the arithmetic series $6n$ . The sum of an arithmetic series is given by 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 $6n$ Part: The first term when $n=0$ is $6\times0 = 0$ , and the last term when $n=68$ is $6\times68 = 408$ . There are $69$ terms in total because we start counting from $n=0$ .
Calculate Constant Term: Now we can calculate the sum of the $6n$ part: $S_{6n} = \frac{69}{2} \times (0 + 408) = \frac{69}{2} \times 408 = 34.5 \times 408 = 14076$ .
Calculate Total Sum: Next, we calculate the sum of the constant term $3$ , which is added $69$ times (once for each value of $n$ from $0$ to $68$ ). This sum is simply $3 \times 69$ .
Calculate Total Sum: Next, we calculate the sum of the constant term $3$ , which is added $69$ times (once for each value of $n$ from $0$ to $68$ ). This sum is simply $3 \times 69$ .Calculating the sum of the constant term: $S_3 = 3 \times 69 = 207$ .
Calculate Total Sum: Next, we calculate the sum of the constant term $3$ , which is added $69$ times (once for each value of $n$ from $0$ to $68$ ). This sum is simply $3 \times 69$ .Calculating the sum of the constant term: $S_3 = 3 \times 69 = 207$ .Finally, we add the two sums together to find the total sum of the series: Total Sum = $S_{6n} + S_3 = 14076 + 207$ .
Calculate Total Sum: Next, we calculate the sum of the constant term $3$ , which is added $69$ times (once for each value of $n$ from $0$ to $68$ ). This sum is simply $3 \times 69$ .Calculating the sum of the constant term: $S_3 = 3 \times 69 = 207$ .Finally, we add the two sums together to find the total sum of the series: Total Sum = $S_{6n} + S_3 = 14076 + 207$ .Adding the two sums gives us the final answer: Total Sum = $14283$ .