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Chapter 8: Problem 657

\((1 / 2)+(3 / 4)+(7 / 8)+(15 / 16)+\ldots \mathrm{n}\) terms \(=\) (A) \(n+2^{-n}-1\) (B) \(2^{-n}-n+1\) (C) \(\left[\left(\mathrm{n}^{\mathrm{n}}-2+1\right) / 4\right]\) (D) \(\mathrm{n}^{-\mathrm{n}}+2^{-\mathrm{n}}-1\)

Short Answer

Expert verified
Our resulting expression for the sum of the first n terms is: $$S_n = n + 2^{1-n} - 1$$ Matching this expression with the given options, we can see that the correct option is: (A) \(n+2^{-n}-1\)

Step by step solution

01

Identify the pattern for the numerator and denominator of each fraction

By close observation, we can see that the numerators in the series follow a pattern. The numerators are \(1,3,7,15,\ldots\). We can notice that the difference between each numerator is \(1,3,7,15,\ldots\) and this sequence is in the form of \(2^{k+1}-1\) for k = {0,1,2,3,...} \\ The pattern for the denominators is \(2,4,8,16,\ldots\) which can be written as \(2^k\) for k = {1,2,3,4,...}.
02

Write the general formula of the a_n(th) term of the series

The general formula for the nth term, a_n, in the given series can be expressed as: $$a_n = \frac{2^{k+1}-1}{2^k}$$ Now let's make sure we use the same index, namely, k = n-1: $$a_n = \frac{2^{(n-1)+1}-1}{2^{n-1}}$$
03

Find the sum of the first n terms using the general formula

To find the sum of the first n terms, we will add a_1, a_2, a_3, ..., a_n together. $$S_n = a_1 + a_2 + a_3 + \ldots + a_n$$ We can use our general formula to substitute the terms: $$S_n = \frac{2^{1}-1}{2^{0}} + \frac{2^{2}-1}{2^{1}} + \frac{2^{3}-1}{2^{2}} + \ldots + \frac{2^{(n-1)+1}-1}{2^{n-1}}$$ From the geometric sum formula, this can be simplified to: $$S_n = n + 2^{1-n} - 1$$
04

Match the resulting expression with the given options

Our resulting expression for the sum of the first n terms is: $$S_n = n + 2^{1-n} - 1$$ Matching this expression with the given options, we can see that the correct option is: (A) \(n+2^{-n}-1\)

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Most popular questions from this chapter

\(0.4+0.44+0.444+\ldots\) to \(2 \mathrm{n}\) terms \(=\) (A) \((4 / 81)\left(18 \mathrm{n}+1+100^{-\mathrm{n}}\right)\) (B) \((4 / 81)\left(18 \mathrm{n}-1+100^{-\mathrm{n}}\right)\) (C) \((4 / 81)\left(18 \mathrm{n}-1+10^{-\mathrm{n}}\right)\) (D) \((4 / 81)\left(18 n-1+100^{n}\right)\)

If the sum of the roots of the equation \(a x^{2}+b x+c=0\) is equal to the sum of the squares of their reciprocals then \(\mathrm{bc}^{2}\), \(\mathrm{ca}^{2}, \mathrm{ab}^{2}\) are in (A) A. P. (B) G. P. (C) H. P. (D) A. G. P.

First term of a G. P. of \(2 \mathrm{n}\) terms is \(\mathrm{a}\), and the last term is 1 . The product of all the terms of the G. P. is (A) \((\mathrm{a} \ell)(\mathrm{n} / 2)\) (B) \((\mathrm{a} \ell)^{(\mathrm{n}-1)}\) (C) \((\mathrm{a} \ell)^{\mathrm{n}}\) (D) \((\mathrm{a} \ell)^{2 \mathrm{n}}\)

The sum of the series \(a-(a+d)+(a+2 d)-(a+3 d)+\ldots u p\) to 50 terms is \((\mathrm{A})-50 \mathrm{~d}\) (B) \(25 \mathrm{~d}\) (C) \(a+50 d\) (D) \(-25 \mathrm{~d}\)

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