Question

Find the sum of the series. make sure you use the formula and show your work for credit.

please help me I didn't get this at all!

Answer 1

The sum of the series is 260.

Explanation:

`\displaystyle \sum^(13)_(n=1)(3n-1)`

The given sigma notation means:

• Sum of the series with nth term (3n - 1), starting with n = 1 and ending with n = 13.

As (3n - 1) is linear, the series is arithmetic.

The first term (n = 1) is:

• a₁ = 3(1) - 1 = 2

The second term (n = 2) is:

• a₂ = 3(2) - 1 = 5

The third term (n = 3) is:

• a₃ = 3(3) - 1 = 8

… and the last term (n = 13) is:

• a₁₃ = 3(13) - 1 = 38

Therefore, we need to find 2 + 5 + 8 + ... + 38.

Since we know that the first term, a, is 2, the last term, l, is 38, and the common difference, d, is 3, we can use the sum of the first n terms formula to calculate the sum of the series of the first 13 terms.

`\begin{aligned}\text{Using:} \quad S_(n)&=(1)/(2)n(a+l)\\\\\implies S_(13)&=(1)/(2)(13)(2+38)\\\\&=(1)/(2)(13)(40)\\\\ &=(520)/(2)\\\\&=260\end{aligned}`

Therefore, the sum of the series is 260.

Answer 2

The sum of the series is 260.

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Each term of given series is expressed as:

• tₙ = 3n - 1

This is an arithmetic progression with 13 terms and the first and last terms are:

• t₁ = 3*1 - 1 = 2
• t₁₃ = 3*13 - 1 = 38

Use the sum of the first terms of AP formula:

• Sₙ = (t₁ + tₙ)*n/2
• S₁₃ = (2 + 38)*13/2 = 40*13/2 = 20*13 = 260