/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 \({ }^{m} C_{r+1}+=\sum_{k=m}^{n... [FREE SOLUTION] | 91Ó°ÊÓ

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\({ }^{m} C_{r+1}+=\sum_{k=m}^{n}{ }^{k} C_{r}=\) (A) \({ }^{n} C_{r+1}\) (B) \({ }^{n+1} C_{r+1}\) (C) \({ }^{n} C_{r}\) (D) None of these

Short Answer

Expert verified
The expression simplifies to \( {}^{n+1} C_{r+1} \), so the answer is (B).

Step by step solution

01

Understanding the Problem

We are given the expression \( {}^{m} C_{r+1} + \sum_{k=m}^{n} {}^{k} C_{r} \) and need to determine which binomial coefficient it simplifies to.
02

Recall Binomial Coefficient Identity

Recall the Pascal's identity in binomial coefficients: \( {}^{k+1} C_{r+1} = {}^{k} C_{r} + {}^{k} C_{r+1} \). This will help simplify the given expression.
03

Simplifying the Expression

Rewrite the given sum using Pascal's identity: \( \sum_{k=m}^{n} {}^{k} C_{r} \) becomes \( {}^{m+1} C_{r+1} + {}^{m+2} C_{r+1} + \ldots + {}^{n+1} C_{r+1} \) by the identity.
04

Adding Extra Term

The term \( {}^{m} C_{r+1} \) in the expression is separated from the sum. By adding the sum terms, we are building \( {}^{n+1} C_{r+1} \). Hence, the whole expression simplifies to a single binomial coefficient.
05

Final Answer

Using the identity, the sum \( {}^{m} C_{r+1} + \sum_{k=m}^{n} {}^{k} C_{r} \) evaluates to \( {}^{n+1} C_{r+1} \). Thus, the correct answer is option (B).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pascal's Identity
Pascal's Identity is a powerful tool in combinatorics that relates to binomial coefficients. It states that:\[ {}^{n+1}C_{k} = {}^nC_{k-1} + {}^nC_k \]Where \( {}^nC_k \) represents the binomial coefficient, or "n choose k," denoting the number of ways to choose k elements from a set of n elements.
This identity is very useful because it provides a way to build a binomial coefficient from two smaller ones. It is visually represented in Pascal's Triangle, where each number is the sum of the two directly above it.
For example, when calculating coefficients for expanding a binomial expression, Pascal's Identity helps create the next row from the known previous one. This makes it essential for simplifying complex problems involving sums of binomial coefficients, as seen in the original exercise where the sum was simplified using this identity.
Combinatorics
Combinatorics is the branch of mathematics focused on counting, arrangement, and combination of objects.
It deals with topics like permutations, combinations, and the principles of counting necessary to solve complex mathematical problems. Understanding combinatorics involves:
  • Recognizing the difference between permutations (where order matters) and combinations (where order does not matter).
  • Using formulas to calculate the number of ways to arrange or select items.
  • Applying these principles to real-world scenarios, such as probability and statistical analysis.
In the context of binomial coefficients, combinatorics allows us to determine the number of ways to select k items from a total of n, represented as \( {}^nC_k \).
This is particularly useful in probability theory, computer science, and decision-making processes, where determining the total possible outcomes or arrangements is crucial. Through practice, students can become proficient in combinatorics, gaining the skills needed to tackle a wide range of mathematical problems.
Mathematical Problem Solving
Mathematical problem solving is about developing strategies to tackle problems efficiently and effectively.
It involves understanding the problem, deriving solutions, and validating answers. Here are key steps in problem-solving:
  • Understanding: Clearly define what the problem asks. Identify knowns, unknowns, constraints, and relevant information.
  • Strategizing: Choose an approach. This might involve breaking the problem into smaller parts, identifying patterns, or using established ideas like Pascal's Identity.
  • Execution: Carry out your plan carefully. Follow logical steps and ensure each leads correctly to the next.
  • Review: Check your results. Look over your work to ensure accuracy and correctness.
In solving the given exercise, these steps guide the process: understanding how to apply combinatorial identities like Pascal's Identity leads to simplifying the expression efficiently. Such practices help students improve their critical thinking and analytical skills, forming a backbone for future mathematical success.

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Most popular questions from this chapter

Instructions: In the following questions an Assertion \((A)\) is given followed by a Reason \((R) .\) Mark your responses from the following options: (A) Assertion(A) is True and Reason(R) is True; Reason(R) is a correct explanation for Assertion(A) (B) Assertion(A) is True, Reason(R) is True; Reason(R) is not a correct explanation for Assertion(A) (C) Assertion(A) is True, Reason(R) is False (D) Assertion(A) is False, Reason(R) is True Assertion: In an examination consisting of 9 papers, a candidate has to pass in more papers than the numbers of papers in which he fails in order to be successful. The number of ways in which he can be unsuccessful 256 . Reason: \({ }^{n} C_{0}+{ }^{n} C_{1}+{ }^{n} C_{2}+\ldots+{ }^{n} C_{n}=2^{n}\)

The number of numbers greater than \(10^{6}\) that can be formed using the digits of the number 2334203, if all the digits of the given number must be used, is (A) 360 (B) 420 (C) 260 (D) None of these

In a certain test there are \(n\) questions. In this test \(2^{k}\) students gave wrong answers to at least \((n-k)\) questions, where \(k=0,1,2, \ldots, n\). If the total number of wrong answers is 4095 , then value of \(n\) is (A) 11 (B) 12 (C) 13 (D) 15

The number of non-negative integral solutions to the system of equations \(x+y+z+u+t=20\) and \(x+y+\) \(z=5\) is (A) 336 (B) 346 (C) 246 (D) None of these

In an examination the maximum marks for each of the three papers are 50 each. Maximum marks for the fourth paper are 100 . The number of ways in which the candidate can score \(60 \%\) marks in aggregate is (A) 110256 (B) 110456 (C) 110556 (D) None of these

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