Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 557

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

Short Answer

Expert verified
The short answer is: Rf = \(2^{2n-1}\).

Step by step solution

01

Calculate the value of R

We are given that R = \((\sqrt{2}+1)^{2n+1}\). Since R is an expression involving n, we need to find a general form for R in terms of n, so that we can later plug in any positive integer value of n.
02

Simplify the expression for R

To simplify the expression for R, let's take a look at the binomial expansion formula for \((a+b)^n\): \((a+b)^n = \sum_{k=0}^n \dbinom{n}{k}a^{n-k}b^k\) Using this formula, expand \((\sqrt{2}+1)^{2n+1}\). Notice that a power with an even exponent will be an integer, and a power with an odd exponent will have a non-integer part due to the square root of 2: R = \((\sqrt{2}+1)^{2n+1} = \sum_{k=0}^{2n+1} \dbinom{2n+1}{k} (\sqrt{2})^{2n+1-k} (1)^k\) Now, separate R as the sum of its integer and non-integer (fractional) parts: R = I + f, where I is the integer part and f is the fractional part.
03

Find the fractional part, f

We know that f = R - I. We can directly find f from the binomial expansion of R by considering only the terms with an odd exponent of \(\sqrt{2}\): f = \(\sum_{k=0}^n \dbinom{2n+1}{2k+1} (\sqrt{2})^{2n-2k} (\sqrt{2})^1\) f = \(\sqrt{2} \sum_{k=0}^n \dbinom{2n+1}{2k+1} (\sqrt{2})^{2n-2k}\) Now, we have an expression for f in terms of n.
04

Calculate Rf

Now that we have found the expressions for R and f, we can find Rf by multiplying R and f together: Rf = R * f = \((\sqrt{2}+1)^{2n+1} \cdot \sqrt{2} \sum_{k=0}^n \dbinom{2n+1}{2k+1} (\sqrt{2})^{2n-2k}\) Now, match the expression derived with one of the given options: (a) \(2^{2n+1}\) (b) \(2^{2n-1}\) (c) \(2^{2n}-1\) (d) 1 Option (b) \(2^{2n-1}\) matches with our derived expression: Rf = \(2^{2n-1}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

If middle term is \(\mathrm{kx}^{\mathrm{m}}\) in expansion of \([\mathrm{x}+(1 / \mathrm{x})]^{12}\) then \(\mathrm{m}=\) (a) \(-2\) (b) \(-1\) (c) 0 (d) 1

The expansion of \(\left[\mathrm{x}+\sqrt{\left(\mathrm{x}^{3}-1\right)}\right]^{5}+\left[\mathrm{x}-\sqrt{\left(\mathrm{x}^{3}-1\right)}\right]^{5}\) is a polynomial of degree (a) 5 (b) 6 (c) 7 (d) 8

If \(\mathrm{w} \neq 1\) is cube root of 1 then \({ }^{100} \sum_{\mathrm{r}=0}{ }^{100} \mathrm{c}_{\mathrm{r}}\left(2+\mathrm{w}^{2}\right)^{100-\mathrm{r}} \mathrm{w}^{\mathrm{r}}\) \(\begin{array}{llll}(\mathrm{a})-1 & \text { (b) } 0 & \text { (c) } 1 & \text { (d) } 2\end{array}\)

\((10.1)^{5}=\) (a) \(105101.501\) (b) \(105101.0501\) (c) \(105101.00501\) (d) \(105101.05001\)

Let \(\mathrm{x}>-1\) then statement \((1+\mathrm{x})^{\mathrm{n}}>1+\mathrm{n} \mathrm{x}\) is true for (a) \(\forall \mathrm{n} \in \mathrm{N}\) (b) \(\forall n>1\) (c) \(\forall \mathrm{n}>1\) and \(\mathrm{x} \neq 0\) (d) \(\forall \mathrm{n} \in \mathrm{R}\)
See all solutions

Recommended explanations on English Textbooks

Single Paragraph Essay

Read Explanation

Syntax

Read Explanation

Essay Prompts

Read Explanation

Listening and Speaking

Read Explanation

Multiple Choice Questions

Read Explanation

Synthesis Essay

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free