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Chapter 8: Problem 592

If the \(1^{\text {st }}\) term and common ratio of a G. P. are 1 and 2 respectively then \(\mathrm{s}_{1}+\mathrm{s}_{3}+\mathrm{s}_{5}+\ldots+\mathrm{s}_{2 \mathrm{n}-1}=\) (A) \((1 / 3)\left(2^{2 n}-5 n+4\right)\) (B) \((1 / 3)\left(2^{2 n+1}-5 n\right)\) (C) \((1 / 3)\left(2^{2 n+1}-3 n-2\right)\) (D) \((1 / 3)\left(2^{2 n+1}-5 n^{2}\right)\)

Short Answer

Expert verified
The sum of all odd-indexed terms in the given geometric progression is \(s_{odd} = (1 / 3)\left(2^{2 n+1}-3 n-2\right)\).

Step by step solution

01

Identify the odd-indexed terms

Since we want to find the sum of the odd-indexed terms, we need to find the general expression for these terms in the G.P. We are given that the first term (\(a_1=1\)) and the common ratio (\(r=2\)). The odd-indexed terms will correspond to the indices \(n=1,3,5,\ldots,2n-1\). We can write the terms with odd indices using the general term formula: \(a_1 = a_1 \times r^{1-1} = 1 \times 2^{0} = 1\) \(a_3 = a_1 \times r^{3-1} = 1 \times 2^{2} = 4\) \(a_5 = a_1 \times r^{5-1} = 1 \times 2^{4} = 16\) \(\cdots\) \(a_{2n-1} = a_1 \times r^{(2n-1)-1} = 1 \times 2^{2n-2}\)
02

Find the sum of all odd-indexed terms

We can now calculate the sum of these odd-indexed terms using the geometric series formula. Recall that for any geometric series, the sum of \(n\) terms, \(S_n = \frac{a_1(r^n - 1)}{r-1}\). We have \(n\) odd-indexed terms, starting at the first term. Since the common ratio is the same, we can also use this formula to find their sum. So, we have: \(s_{odd} = S_{2n-1} = \frac{1(2^{(2n-2)+1}-1)}{2-1} = 2^{2n-1}-1\)
03

Compare with given options

We are given four options. Now, let's compare our obtained sum with each option: (A) \((1 / 3)\left(2^{2 n}-5 n+4\right)\neq 2^{2n-1}-1\) (B) \((1 / 3)\left(2^{2 n+1}-5 n\right)\neq 2^{2n-1}-1\) (C) \((1 / 3)\left(2^{2 n+1}-3 n-2\right) = 2^{2n-1}-1\) (D) \((1 / 3)\left(2^{2 n+1}-5 n^{2}\right)\neq 2^{2n-1}-1\) From our comparison, we can see that the correct option is (C). Therefore, the sum of the odd-indexed terms in this geometric progression is given by: \(s_{odd} = (1 / 3)\left(2^{2 n+1}-3 n-2\right)\)

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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) \((4 / 81)\left(18 n-1+100^{n}\right)\)

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