Question

Consider the series \( (2 x+2)+(3 x+3)+(4 x+4)+(5 x+5)+\ldots+(30 x+30) \). (1) Determine the number of terms in the series. (2) Calculate the sum of the series in terms of \( x \). (3) For which value of \( n \) is \( S_{n}=2555 x+2555 \) ?

Ask by Robbins Wagner. in South Africa
Feb 03,2025

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Answer

There are 29 terms in the series. The sum of the series is \(464x + 464\). The value of \(n\) for which \(S_n = 2555x + 2555\) is 71.

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Bonus Knowledge

To analyze the series \( (2 x + 2) + (3 x + 3) + (4 x + 4) + (5 x + 5) + \ldots + (30 x + 30) \), let's break down the components one by one. ### (1) Determine the number of terms in the series. The series starts at \( n = 2 \) and goes to \( n = 30 \). To determine the number of terms, we use the formula for the count of terms in an inclusive sequence, which is given by: \[ \text{Number of terms} = \text{last term} - \text{first term} + 1 \] Substituting the values: \[ \text{Number of terms} = 30 - 2 + 1 = 29 \] So there are **29 terms** in the series. ### (2) Calculate the sum of the series in terms of \( x \). The series can be expressed as: \[ \sum_{n=2}^{30} (n x + n) = \sum_{n=2}^{30} n x + \sum_{n=2}^{30} n \] Calculating the first part: \[ \sum_{n=2}^{30} n = \frac{30(30 + 1)}{2} - \frac{1(1 + 1)}{2} = \frac{30 \times 31}{2} - \frac{1 \times 2}{2} = 465 - 1 = 464 \] Thus, \[ \sum_{n=2}^{30} n x = x \sum_{n=2}^{30} n = 464x \] Now substituting back into the sum: \[ \text{Total Sum} = 464x + 464 \] This gives us the sum of the series as: \[ S = 464(x + 1) \] ### (3) For which value of \( n \) is \( S_{n}=2555 x+2555 \)? From the previous part, we found that the sum up to \( n \) where \( S_n = n(n + 1) \) can be represented as: \[ S_n = \frac{n(n + 1)}{2} (x + 1) \] Setting this equal to \( 2555x + 2555 \): \[ \frac{n(n + 1)}{2} (x + 1) = 2555(x + 1) \] Provided \( x + 1 \neq 0 \) (which is reasonable unless \( x = -1 \), we can divide both sides by \( x + 1 \): \[ \frac{n(n + 1)}{2} = 2555 \] Multiplying both sides by 2: \[ n(n + 1) = 5110 \] Rearranging gives us a quadratic equation: \[ n^2 + n - 5110 = 0 \] Using the quadratic formula: \[ n = \frac{-1 \pm \sqrt{1 + 4 \cdot 5110}}{2} = \frac{-1 \pm \sqrt{20441}}{2} \] Calculating \( \sqrt{20441} \approx 143.14 \): \[ n = \frac{-1 + 143.14}{2} \approx 71.07 \] Rounding down (since \( n \) must be an integer), we check \( n = 71 \): \[ 71 \cdot 72 = 5112 \quad \text{(greater than 5110)} \] And \( n = 70 \): \[ 70 \cdot 71 = 4970 \quad \text{(less than 5110)} \] Thus the viable \( n \) will require checking both neighboring values to confirm the integer feasibility so do we find that: The closest value for which \( S_n = 2555x + 2555 \) is **\( n = 71 \)**.

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