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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

The sum of \(\mathrm{n}\) terms of the series \(12+16+24+40+\ldots\) is (A) \(8\left(2^{\mathrm{n}}-1\right)+8 \mathrm{n}\) (B) \(4\left(2^{\mathrm{n}}-1\right)+8 \mathrm{n}\) (C) \(2\left(2^{n}-1\right)+6 n\) (D) \(3\left(2^{\mathrm{n}}-1\right)+8 \mathrm{n}\)

The greatest value of \(n\) for which \(1+(1 / 2)+\left(1 / 2^{2}\right)+\ldots\) \(+\left(1 / 2^{\mathrm{n}}\right)<2\) is \((\mathrm{n} \in \mathrm{N})\) (A) 100 (B) 10 (C) 1000 (D) none of these

If \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are in \(\mathrm{A} . \mathrm{P}\). and geometric means of ac and \(\mathrm{ab}, \mathrm{ab}\) and \(b c\), ba nad \(c b\) are \(d\), e, f respectively then \(d^{2}, e^{2}, f^{2}\) are in (A) A. P. (B) G. P. (C) H. P. (D) A. G P.

If the harmonic mean and geometric mean of two numbers a and \(b\) are 4 and \(3 \sqrt{2}\) respectively then the interval \([a, b]=\) (A) \([3,6]\) (B) \([2,7]\) (C) \([4,5]\) (D) \([1,8]\)

Find out three numbers which are in G. P. such that their summation is 13 and the sum of their squares is 91 (A) \(3,1,9\) (B) \(1,3,9\) (C) \(1,9,3\) (D) \((13 / 3),(13 / 3),(13 / 3)\)
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