Chapter 7: Problem 556

\(\mathrm{R}=(3+\sqrt{5})^{2 \mathrm{n}}\) and \(\mathrm{f}=\mathrm{R}-[\mathrm{R}]\), Where [] is an integer part function then \(\mathrm{R}(1-\mathrm{f})=\) (a) \(2^{2 \mathrm{n}}\) (b) \(4^{2 \mathrm{n}}\) (c) \(8^{2 \mathrm{n}}\) (d) \(1^{2 \mathrm{n}}\)

Short Answer

Expert verified

The correct answer for \(\mathrm{R}(1-\mathrm{f})\) is (d) \(1^{2 \mathrm{n}}\).

Step by step solution

Express f as a subtraction between R and integer part of R

We are given the expression for \(\mathrm{f}=\mathrm{R}-[\mathrm{R}]\) where \([\mathrm{R}]\) represents the integer part of \(\mathrm{R}\). We will use this expression later when computing \(\mathrm{R}(1-\mathrm{f})\).

Find 1-f

We need to compute \(1-\mathrm{f}\). Using the expression for \(\mathrm{f}\) we found earlier, we can rewrite this as: \(1-\mathrm{f} = 1 - (\mathrm{R}-[\mathrm{R}])\). We can then simplify this as: \(1-\mathrm{f}=1-\mathrm{R}+[\mathrm{R}]=[\mathrm{R}]-\mathrm{R}+1\)

Calculate R(1-f)

Now that we have the expression for \(1-\mathrm{f}\), we can compute \(\mathrm{R}(1-\mathrm{f})\). We can substitute the expression for \([\mathrm{R}] - \mathrm{R} + 1\) that we found in Step 2: \(\mathrm{R}(1-\mathrm{f})=\mathrm{R}([\mathrm{R}]-\mathrm{R}+1)\) \((3+\sqrt{5})^{2n}([\mathrm{R}]-\mathrm{R}+1)\)

Notice the Binomial Theorem Pattern

Since \(\mathrm{R}\) has the form of \((3+\sqrt{5})^{2n}\), it will expand using the binomial theorem. Pay attention to the even powers of the expansion, as odd powers will include irrational numbers. We want to understand the relationship between the integer part of \(\mathrm{R}\) and the irrational part of \(\mathrm{R}\). Consider the expansion of \((3+\sqrt{5})^2\). The even terms are: \((3+\sqrt{5})^2= 3^2 + 2(3)(\sqrt{5}) + 5= 9 + 6\sqrt{5} + 5= 14 + 6\sqrt{5}\). We can conclude that the integer part of \((3+\sqrt{5})^2\) is 14 and the irrational part is \(6\sqrt{5}\).

Compute R(1-f) and Match the Answer

From the reasoning in Step 4, we can see that \(\mathrm{R}\) will generally have terms containing even powers of 3 and 5 (which will be integer) and the irrational parts containing some multiples of \(\sqrt{5}\). Thus, the term \([\mathrm{R}]-\mathrm{R} +1\) will have an irrational part that cancels out the irrational part of \(\mathrm{R}\). Thus, when looking at the expanded form of \(\mathrm{R}(1-\mathrm{f})\), only the even powers of the expansion and the integer coefficients matter. Since the constant term in the expansion is 1, our \(\mathrm{R}(1-\mathrm{f})\) has the form of \(1^{2n}\). Thus, the correct answer is: (d) \(1^{2 \mathrm{n}}\)

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\(\mathrm{R}=(3+\sqrt{5})^{2 \mathrm{n}}\) and \(\mathrm{f}=\mathrm{R}-[\mathrm{R}]\), Where [] is an integer part function then \(\mathrm{R}(1-\mathrm{f})=\) (a) \(2^{2 \mathrm{n}}\) (b) \(4^{2 \mathrm{n}}\) (c) \(8^{2 \mathrm{n}}\) (d) \(1^{2 \mathrm{n}}\)

Short Answer

Step by step solution

Express f as a subtraction between R and integer part of R

Find 1-f

Calculate R(1-f)

Notice the Binomial Theorem Pattern

Compute R(1-f) and Match the Answer

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