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Chapter 7: Problem 558

\(\left[5^{(1 / 2)}+7^{(1 / 8)}\right]^{1024}\) has number of rational terms \(=\) (a) 0 (b) 129 (c) 229 (d) 178

Short Answer

Expert verified
The number of rational terms in the expression \(\left[5^{(1 / 2)}+7^{(1 / 8)}\right]^{1024}\) is 129. Answer: (b).

Step by step solution

01

Understand the binomial theorem

The binomial theorem states that for any non-negative integer n and any real numbers a and b: \((a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k\) Where \(\binom{n}{k}\) is the binomial coefficient and can be calculated using the formula: \(\binom{n}{k} = \frac{n!}{k! (n-k)!}\).
02

Apply the binomial theorem to the given expression

We will apply the binomial theorem to the expression \(\left[5^{(1 / 2)}+7^{(1 / 8)}\right]^{1024}\) to expand it. Let's set \(a = 5^{(1 / 2)}\) and \(b = 7^{(1 / 8)}\). We get: \[(5^{(1 / 2)} + 7^{(1 / 8)})^{1024} = \sum_{k=0}^{1024} \binom{1024}{k} (5^{(1 / 2)})^{1024 - k} (7^{(1 / 8)})^k\]
03

Determine when the terms are rational

For a term to be rational, the exponents of both 5 and 7 must be integers. Let's analyze the exponent of 5, which is \((1024 - k) \cdot \frac{1}{2}\). This exponent will be an integer when \(1024 - k\) is even. Similarly, the exponent of 7, which is \(k \cdot \frac{1}{8}\), will be an integer when k is a multiple of 8.
04

Find all possible values of k

For a term to be rational, \(1024 - k\) must be even, and k must be a multiple of 8. Let's find all possible values of k that satisfy these conditions. Since \(1024 - k\) must be even, k can be any even number between 0 and 1024. And because k must also be a multiple of 8, k can be any multiple of 8 between 0 and 1024.
05

Count the number of possible values for k

To find the number of possible values for k, we can divide the difference between the largest and smallest possible values for k by their common difference, and add 1. That is: \[\frac{1024 - 0}{8} + 1 = \frac{1024}{8} + 1 = 128 + 1 = 129\] So, there are 129 possible values of k.
06

Conclusion

There are 129 possible values of k such that the terms in the expression \(\left[5^{(1 / 2)}+7^{(1 / 8)}\right]^{1024}\) are rational. Therefore, the answer is: (a) 0 (b) 129 (c) 229 (d) 178 The correct answer is (b) 129.

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Most popular questions from this chapter

\(3^{\mathrm{rd}}\) term in expansion of \(\left[(1 / \mathrm{x})+\mathrm{x}^{(\log )}{ }_{10}(\mathrm{x})\right]^{5}\) is 1000 then \(\mathrm{x}=\) (a) 10 (b) 100 (c) 1000 (d) None

It coefficients of middle terms in expansion of \((1+\lambda \mathrm{x})^{8}\) and \((1-\lambda \mathrm{x})^{6}\) are equal then \(\lambda=\) (a) \((2 / 7)\) (b) \([(-2) / 7]\) (c) \([(-3) / 7]\) (d) None of these

th term is constant term in expansion of \(\left[\left(3 / x^{2}\right)+(\sqrt{x} / 3)\right]^{10}, x \neq 0\) (a) 4 (b) 7 (c) 8 (d) 9

Constant term in expansion of \(\left[\left\\{(\mathrm{x}+1) /\left\\{(\mathrm{x})^{(2 / 3)}-(\mathrm{x})^{(1 / 3)}+1\right\\}\right\\}-\left\\{(\mathrm{x}-1) /\left\\{\left(\mathrm{x}-(\mathrm{x})^{(1 / 2)}\right\\}\right\\}\right]^{10}\right.\) is \(\ldots\) (a) 190 (b) 200 (c) 210 (d) 220

Constant term in expansion of \([1+(2 / \mathrm{x})-(2 / \mathrm{x})]^{4}\) is ........ (a) 5 (b) \(-5\) (c) 4 (d) \(-4\)
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