Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 498

In the expansion of \((\mathrm{x}+\mathrm{y})^{13}\) the co-efficient of \(3 \mathrm{rd}\) term and th terms are equal. (a) 12 (b) 11 (c) 8 (d) 13

Short Answer

Expert verified
None of the given options match the obtained value of 'n', which is 5. There might be a mistake in the given options, or the given data in the question might require a revision.

Step by step solution

01

Find the formula for coefficients using the binomial theorem

According to the binomial theorem, the general term in the expansion of \((x+y)^n\) is given by: \[T_r = \binom{n}{r-1}x^{n-r+1}y^{r-1}\] where \(T_r\) represents the \(r^{th}\) term in the expansion.
02

Find the 3rd and 4th terms in the expansion

Given that the coefficients of the 3rd and 4th terms are equal, we need to find the coefficients of these terms using the binomial theorem. For the 3rd term, \(r=3\): \[T_3 = \binom{n}{2}x^{n-1}y^2\] For the 4th term, \(r=4\): \[T_4 = \binom{n}{3}x^{n-2}y^3\]
03

Equate the coefficients of the 3rd and 4th terms

Since the coefficients of the 3rd and 4th terms are equal, we need to equate the coefficients and solve for 'n': \[\binom{n}{2} = \binom{n}{3}\]
04

Solve for 'n'

Using the combination formula, we can rewrite the equation as: \[\frac{n!}{2!(n-2)!} = \frac{n!}{3!(n-3)!}\] Now, we can simplify and solve for 'n': \[\frac{1}{2(n-2)} = \frac{1}{3(n-3)}\] Cross-multiplying, we get: \[3(n-3) = 2(n-2)\] Expanding and solving for 'n', we get: \[3n - 9 = 2n - 4\] \[n = 5\] Given the options (a) 12, (b) 11, (c) 8, and (d) 13, none of them match with the obtained value of 'n'. There might be a mistake in the given options, or the given data in the question might require a revision.

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

The sum of coefficient of middle terms in expansion of \((1+x)^{2 n-1}\) is (a) \({ }^{(2 \mathrm{n}-1)} \mathrm{C}_{\mathrm{n}}\) (b) \({ }^{(2 \mathrm{n}-1)} \mathrm{C}_{(\mathrm{n}-1)}\) (c) \({ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{n}}\) (d) \({ }^{2 \mathrm{n}} \mathrm{C}_{(\mathrm{n}+1)}\)

If co-efficient of \((\mathrm{r}+2)\) th term and \(3 \mathrm{r}\) th term are equal in expansion of \((1+\mathrm{x})^{2 \mathrm{n}}, \mathrm{n}, \mathrm{r} \in \mathrm{N}, \mathrm{r}>1, \mathrm{n}>2\) then \(\mathrm{n}=\) (a) \(3 \mathrm{r}\) (b) \(3 \mathrm{r}+1\) (c) \(2 \mathrm{r}\) (d) \(2 r+1\)

\(\mathrm{s}(\mathrm{k}): 1+3+5+\ldots+(2 \mathrm{k}-1)=3+\mathrm{k}^{2}\) then which statement is true ? (a) \(\mathrm{s}(\mathrm{k}) \Rightarrow \mathrm{s}(\mathrm{k}+1)\) (b) \(\mathrm{s}(\mathrm{k}) \Rightarrow \mathrm{s}(\mathrm{k}+1)\) (c) s (1) is true (d) Result is proved by Principle of Mathematical induction

Co-efficient of \(\mathrm{x}^{5}\) in expansion of \((1+2 \mathrm{x})^{6}(1-\mathrm{x})^{7}\) is....... (a) 150 (b) 171 (c) 192 (d) 161

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

Prosody

Read Explanation

International English

Read Explanation

History of English Language

Read Explanation

Creative Story

Read Explanation

Rhetorical Analysis Essay

Read Explanation

Blog

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