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Chapter 8: Problem 663

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

Short Answer

Expert verified
The greatest value of \(n\) for which the inequality holds is \(n=100\). The answer is (A) 100.

Step by step solution

01

(Step 1: Identify the relevant formula for the geometric series)

For a geometric series, sum(\(S\)) can be represented as: \(S = \frac{a(1 - r^n)}{1 - r}\), where \(a\) is the first term, \(r\) is the common ratio and \(n\) is the number of terms. In our given series, \(a = 1\) and \(r = \frac{1}{2}\), so we'll substitute these values into the formula and use the given inequality condition.
02

(Step 2: Apply the inequality condition to the summation formula)

According to the exercise, \(\displaystyle\sum_{k=0}^n \frac{1}{2^k} < 2\). Substitute the values of \(a\) and \(r\) into the formula: \(\frac{1(1 - (\frac{1}{2})^n)}{1 - \frac{1}{2}} < 2\)
03

(Step 3: Simplify and solve for n)

Solve the inequality and find the value of \(n\): \(\frac{1 - (1/2)^n}{1/2} < 2\) Multiply both sides by 1/2: \(1 - (1/2)^n < 1\) Move terms to the left side of the inequality: \((1/2)^n > 0\) Since any value of \(n\) will result in a positive value for \((1/2)^n\), we need to find the greatest value possible for \(n\) which still satisfies the initial inequality: Subtract terms on the left-hand side: \((1/2)^n = 1 - \frac{1}{2^n}\) At this point, we know that the largest possible value of \(n\) that satisfies our inequality is when \((1/2)^n\) approaches zero (but is just slightly greater).
04

\(Step 4: Observe which option best satisfies the inequality and choose the answer\)

Given the available options, let's check which one satisfies the inequality: (A) 100: \(1/2^{100} \approx 7.9 \times 10^{-31}\) (Very close to zero) (B) 10: \(1/2^{10} \approx 9.8 \times 10^{-4}\) (C) 1000: \(1/2^{1000}\) (Even closer to zero, but not one of the options) (D) None of these Considering the values, we can see that the option (C) 1000 will result in a value of \((1/2)^n\) that is even closer to zero, which is not provided in the given options. Therefore, the greatest given value of \(n\) that satisfies the inequality is option (A) 100.

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Most popular questions from this chapter

\(1+5+14+30+\ldots \mathrm{n}\) terms \(=\) (A) \([\\{(n+2)(n+3)\\} /(12)]\) (B) \([\\{n(n+1)(n+5)\\} /(12)]\) (C) \([\\{n(n+2)(n+3)\\} /(12)]\) (D) \(\left[\left\\{n(n+1)^{2}(n+2)\right\\} /(12)\right]\)

If for the triangle whose perimeter is \(37 \mathrm{cms}\) and length of sides are in G. P. also the length of the smallest side is \(9 \mathrm{cms}\) then length of remaining two sides are and (A) 12,16 (B) 14,14 (C) 10,18 (D) 15,13

\(0.4+0.44+0.444+\ldots\) 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 first 101 terms of an A. P. is 0 and If 1 be the first term of the A. P. then the sum of next 100 terms is \((\mathrm{A})-101\) (B) 201 (C) \(-201\) (D) \(-200\)

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.
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