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Chapter 7: Problem 584

Numbers of rational terms in expansion \(\left.\left[3^{(1 / 2)}+5^{(1 / 8)}\right)\right]^{256}\) are (a) 33 (b) 34 (c) 35 (d) 32

Short Answer

Expert verified
The number of rational terms in the expansion \(\left[3^{\frac{1}{2}} + 5^{\frac{1}{8}}\right]^{256}\) is 33.

Step by step solution

01

Recall the binomial theorem

Recall that according to the binomial theorem, for any positive integer n and any real numbers x and y, the expansion of (x+y)^n is given by: \((x+y)^n = \sum_{k=0}^{n}\binom{n}{k}x^{n-k}y^k\) In our case, x = \(3^{\frac{1}{2}}\) and y = \(5^{\frac{1}{8}}\), with n = 256.
02

Identify rational term conditions

For a term in the expansion to be rational, both the base 3 and base 5 components need to have rational exponents. That is, \(3^{\frac{1}{2}(256-k)}\) must be a rational number, and \(5^{\frac{1}{8}k}\) must be a rational number.
03

Determine exponents required for rational numbers

Looking at the base 3 exponent: \(\frac{1}{2}(256-k) = 128 - \frac{1}{2}k\) For the expression to result in a rational number, the exponent of 3 must be an integer. Similarly, looking at the base 5 exponent: \(\frac{1}{8}k\) For the expression to result in a rational number, the exponent of 5 must be an integer.
04

Combine the conditions and solve for k

Combining the conditions, we have: \(128 - \frac{1}{2}k\) must be an integer, \(\frac{1}{8}k\) must be an integer. Now, notice that if \(\frac{1}{8}k\) must be an integer, k must be a multiple of 8. Let \(k = 8m\), where m is an integer. Now we can rewrite the base 3 exponent condition: \(128 - \frac{1}{2}(8m)\) \(128 - 4m\) This way, the integer m can take any value ranging from 0 to 32.
05

Count the number of rational terms

As m ranges from 0 to 32, we get a total of 33 values for m, and thus 33 values for k that produce rational terms in the expansion. So the number of rational terms in the given expression is 33. #Answer# (a) 33

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Most popular questions from this chapter

If rth term in the expansion of \(\left[a x^{p}+\left(b / x^{q}\right)\right]^{n}\) is constant, then \(\mathrm{r}=\) (a) \([(q n) \bar{m}+q)]+1\) (b) \([(\mathrm{pn}) /(\mathrm{p}-\mathrm{q})]+1\) (c) \([(\mathrm{pn}) /(\mathrm{p}+\mathrm{q})]+1\) (d) \([(\mathrm{pn}) /(\mathrm{p}-\mathrm{q})]-1\)

In the expansion of \(\left[x^{3}-\left(1 / x^{2}\right)\right]^{15}\), the constant term is (a) \(-{ }^{15} \mathrm{C}_{9}\) (b) \({ }^{15} \mathrm{C}_{9}\) (b) 0 (d) \({ }^{15} \mathrm{C}_{11}\)

It \(\mathrm{A}\) and \(\mathrm{B}\) are coefficients of \(\mathrm{x}^{\mathrm{r}}\) and \(\mathrm{x}^{\mathrm{n}-\mathrm{r}}\) respectively in expansion of \((1+x)^{n}\) then \(=\) (a) \(\mathrm{A}+\mathrm{B}=\mathrm{n}\) (b) \(\mathrm{A}=\mathrm{B}\) (c) \(\mathrm{A}+\mathrm{B}=2^{\mathrm{n}}\) (d) \(A-B=2^{n}\)

The interval in which \(\mathrm{x}(>0)\) must lie so that greatest term in the expansion of \((1+\mathrm{x})^{2 \mathrm{n}}\) has the greatest coefficient is (a) \([\\{(n-1) / n\\},\\{n /(n-1)\\}]\) (b) \([\\{\mathrm{n} /(\mathrm{n}+1)\\},\\{(\mathrm{n}+1) / \mathrm{n}\\}]\) (b) \([\\{n /(n+2)\\},\\{(n+2) / n\\}]\) (d) None

\({ }^{\mathrm{n}} \mathrm{C}_{1}+2 \cdot{ }^{\mathrm{n}} \mathrm{C}_{2}+3 \cdot{ }^{\mathrm{n}} \mathrm{C}_{3}+\ldots+\mathrm{n} \cdot{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}=\) (a) n \(\cdot 2^{n-1}\) (b) \((\mathrm{n}-1) 2^{\mathrm{n}-1}\) (c) \((\mathrm{n}+1) 2^{\mathrm{n}-1}\) (d) \((\mathrm{n}-1) 2^{\mathrm{n}}\)
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