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Chapter 8: Problem 612

\(2+12+36+80+\ldots \mathrm{n}\) terms \(=\) (A) \([\\{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)(3 \mathrm{n}+5) \overline{\\} /(24})]\) (B) \([\\{n(n+1)(n+2)(3 n+1)\\} /(12)]\) (C) \([\\{n(n+1)(n+3)(n+5)\\} /(24)]\) (D) \([\\{n(n+1)(n+2)(n+3)\\} /(12)]\)

Short Answer

Expert verified
The short answer is: \(S_n = \frac{n(n+1)(n+2)(3n+1)}{12}\) (Option B).

Step by step solution

01

Identify the sequence pattern

First, let's look at the given sequence carefully: 2, 12, 36, 80, ... Notice the differences between consecutive terms: 12 - 2 = 10 36 - 12 = 24 80 - 36 = 44 The differences are increasing by 14 each time (24 - 10 = 14 and 44 - 24 = 14). Now, let's rewrite the differences as product of two consecutive integers: 10 = 2 × 5 24 = 3 × 8 44 = 4 × 11 We can now see a pattern: the first number in the product is just the term number, n (starting from 2), and the second number is n × 2 + 3. That means we can write a general term for the sequence as: \(a_n = n(n × 2 + 3) = 2n^2 + 3n\)
02

Find the sum formula for the first n terms

Now, we need to find the sum of the first n terms of the sequence. We'll call this sum S_n. Using the formula for the sum of the first n terms of a quadratic sequence, we can write: \(S_n = \frac{n}{6} [2(a_1 + a_n) + (n - 1)d]\) Where: - n is the number of terms - a_1 is the first term - a_n is the last term - d is the common difference Substitute our values for a_n and the common difference, d: \(a_1 = 2(1)^2 + 3(1) = 5\) \(a_n = 2n^2 + 3n\) \(d = 4n - 2\) Plugging these into the sum formula: \(S_n = \frac{n}{6} [2(5 + (2n^2 + 3n)) + (n - 1)(4n - 2)]\)
03

Simplify the sum formula and match with the given options

Now, we will simplify the sum expression and check which option it matches: \(S_n = \frac{n}{6} [2(5 + 2n^2 + 3n) + (4n^2 - 2n)]\) \(S_n = \frac{n}{6} [10 + 4n^2 + 6n + 4n^2 - 2n]\) \(S_n = \frac{n}{6} [8n^2 + 4n + 10]\) \(S_n = \frac{n(2n+1)(n+1)(n+2)}{6}\) Now it's clearer that this sum formula fits option (B): \(S_n = [n(n+1)(n+2)(3n+1)/(12)]\) So the correct answer is option (B).

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Most popular questions from this chapter

\(\left(1^{3} / 1\right)+\left[\left(1^{3}+2^{3}\right) /(1+2)\right]+\left[\left(1^{3}+2^{3}+3^{3}\right) /(1+2+3)\right]\) \(+\ldots 15\) terms (A) 446 (B) 680 (C) 600 (D) 540

If sec \((\mathrm{x}-\mathrm{y}), \sec \mathrm{x}\) and \(\mathrm{sec}(\mathrm{x}+\mathrm{y})\) are in A. P., then \(\cos x \sec (y / 2)=\ldots \ldots \ldots(y \neq 2 n \pi, n \in I)\) \(\begin{array}{llll}\text { (A) } \pm \sqrt{2} & \text { (B) } \pm(1 / \sqrt{2}) & \text { (C) } \pm 2 & \text { (D) } \pm(1 / 2)\end{array}\)

If \((1 / a),(1 / b),(1 / c)\) are in A. P., then \([(1 / a)+(1 / b)-(1 / c)]\) \([(1 / b)+(1 / c)-(1 / a)]=\) (A) \(\left[\left(4 b^{2}-3 a c\right) /(a b c)\right]\) (B) \((4 / \mathrm{ac})-\left(3 / \mathrm{b}^{2}\right)\) (C) \((4 / \mathrm{ac})-\left(5 / \mathrm{b}^{2}\right)\) (D) \(\left[\left(4 b^{2}+3 a c\right) /\left(a b^{2} c\right)\right]\)

If a set \(\mathrm{A}=\\{3,7,11, \ldots, 407\\}\) and a set \(\mathrm{B}=\\{2,9,16, \ldots, 709\\}\) then \(\mathrm{n}(\mathrm{A} \cap \mathrm{B})=\) (A) 13 (B) 14 (C) 15 (D) 16

Find \(\mathrm{a}, \mathrm{b}\) and \(\mathrm{c}\) between 2 and 18 such that \(\mathrm{a}+\mathrm{b}+\mathrm{c}=25\), \(2, a, b\) are in A. P. and \(b, c, 18\) are in G. P. (A) \(5,8,12\) (B) \(4,8,13\) (C) \(3,9,13\) (D) \(5,9,11\)
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