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Chapter 8: Problem 659

The sum of the series \((3 / 4)+(5 / 36)+[7 /(144)], \ldots\) up to 11 terms is (A) \([(120) \overline{/(121)]}\) (B) \([(143) /(144)]\) (C) 1 (A) \([(144) /(143)]\)

Short Answer

Expert verified
The correct answer is (A) \(\frac{121}{22}\).

Step by step solution

01

Find the general expression for the nth term of the series.

We can see that the numerators of each term are consecutive odd numbers and the denominators are consecutive perfect squares multiplied by 4. Therefore, the nth term of the series can be written as: T_n = \(\frac{(2n + 1)}{4n^2}\)
02

Use the sum formula for arithmetic progressions.

The sum of an arithmetic progression up to n terms can be calculated using the formula: S_n = n * (a_1 + T_n) / 2 In this case, a_1 is the 1st term, which is 3/4, and T_n is the nth term, discussed in Step 1. We need to sum up to 11 terms, so n = 11.
03

Substitute the values and simplify.

Plug the values into the formula and simplify: S_11 = 11 * \(\frac{(3/4) + (2 * 11 + 1) / (4 * 11^2)}\) Simplify the expression inside the parentheses: S_11 = 11 * \(\frac{(3/4) + (23) / (484)}\) Now, find a common denominator for both the fractions inside the parentheses and then add them: S_11 = 11 * \(\frac{(363 + 23) / (484)}\) S_11 = 11 * \(\frac{386 / 484}\) Simplify the fraction: S_11 = 11 * \(\frac{193 / 242}\) Now, multiply the fraction by 11: S_11 = \(\frac{193 * 11}{242}\) S_11 = \(\frac{2123}{242}\) Finally, compact the fraction to get the final answer: S_11 = \(\frac{121}{22}\) So, the correct answer is (A) \(\frac{121}{22}\).

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Most popular questions from this chapter

Sum of products of first n natural numbers taken two at a time is (A) \(\left[\left\\{\overline{ \left.n\left(n^{2}-1\right)(3 n+2)\right\\} /(24)}\right]\right.\) (B) \(\left[\left\\{n(n+1)^{2}(3 n+2)\right\\} /(72)\right]\) (C) \(\left[\left\\{n^{2}(n+1)(3 n+2)\right\\} /(48)\right]\) (D) \([\\{n(n+1)(n+2)(3 n+2)\\} /(96)]\)

\(\left(3 / 1^{2}\right)+\left[5 /\left(1^{2}+2^{2}\right)\right]+\left[7 /\left(1^{2}+2^{2}+3^{2}\right)\right] \ldots\) upto n terms \(-\) (A) \(\left[\left(6 \mathrm{n}^{2}\right) /(\mathrm{n}+1)\right]\) (B) \([(6 n) /(\mathrm{n}+1)]\) (C) \([\\{6(2 \mathrm{n}-1)\\} /(\mathrm{n}+1)]\) (D) \(\left[\left\\{3\left(n^{2}+1\right)\right\\} /(n+1)\right]\)

A. \(\mathrm{M}\) of the three numbers which are in G. P. is \([(14) / 3]\) If adding 1 in first and second number and subtracting 1 from the third number, resulting numbers are in \(\mathrm{A} . \mathrm{P}\). then the sum of the squares of original three numbers is (A) 91 (B) 80 (C) 84 (D) 88

If the sides of a \(\triangle \mathrm{ABC}\) are in A. P. and the greatest angle is double the smallest. The ratio of the sides of \(\triangle \mathrm{ABC}\) is (A) \(3: 4: 5\) (B) \(5: 12: 13\) (C) \(4: 5: 6\) (D) \(5: 6: 7\)

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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