Problem 75
Question
\({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}+{ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{n}}+{ }^{\mathrm{n}+2} \mathrm{C}_{\mathrm{n}}+\ldots+{ }^{\mathrm{n}+\mathrm{k}} \mathrm{C}_{\mathrm{n}}=\) \((\text { A })^{n+k-1} C\) (B) \({ }^{n+k} C_{n+1}\) (C) \({ }^{\mathrm{n}+\mathrm{k}+{ }^{1}} C_{\mathrm{n}+1}\) (D) none of these
Step-by-Step Solution
Verified Answer
The answer is Option (C) \(^{n+k+1}C_{n+1}\).
1Step 1: Understand the Expression
We need to solve the given binomial coefficient sum: \[ { }^{n}C_{n} + { }^{n+1}C_{n} + { }^{n+2}C_{n} + \ldots + { }^{n+k}C_{n} \].This is an expression representing a sequence of binomial coefficients starting with choosing \( n \) items from \( n \) and ending with choosing \( n \) items from \( n+k \).
2Step 2: Recognize the Pattern
The expression can be summed up using the hockey stick identity (aka Christmas Stocking identity), which states:\[ { }^{r}C_{r} + { }^{r+1}C_{r} + { }^{r+2}C_{r} + \ldots + { }^{r+s}C_{r} = { }^{r+s+1}C_{r+1} \].Here, \( r = n \) and the last term index is \( r+s = n+k \).
3Step 3: Apply the Hockey Stick Identity
Using the identity \[ { }^{n}C_{n} + { }^{n+1}C_{n} + { }^{n+2}C_{n} + \ldots + { }^{n+k}C_{n} = { }^{n+k+1}C_{n+1} \],we apply it directly to obtain the result of the original sum.
4Step 4: Find the Matching Option
From the identity result, we have \[{ }^{n+k+1}C_{n+1} \].Referring to the options given, Option (C) \(^{n+k+1}C_{n+1}\) correlates with our result. Thus, it is the correct answer.
Key Concepts
Hockey Stick IdentityCombinatorial IdentitiesBinomial Theorem
Hockey Stick Identity
The Hockey Stick Identity, also known as the Christmas Stocking Identity, is a fascinating pattern found in binomial coefficients. Imagine a Pascal's Triangle, an arrangement of numbers in a triangular format where each number is the sum of the two numbers directly above it.
The identity is named for its hockey stick appearance when the numbers are selected, starting from an element and moving diagonally down the triangle, and then making a turn and summing vertically. The specific identity is given by:\[{ }^{r}C_{r} + { }^{r+1}C_{r} + { }^{r+2}C_{r} + \ldots + { }^{r+s}C_{r} = { }^{r+s+1}C_{r+1}\]
Here, these represent binomial coefficients. The sum starts at \(^{r}C_{r}\) and continues to \(^{r+s}C_{r}\), resulting in the sum \(^{r+s+1}C_{r+1}\). It simplifies evaluating series of coefficients efficiently by identifying these patterns. This identity is crucial in problems dealing with sequences of binomial coefficients.
The identity is named for its hockey stick appearance when the numbers are selected, starting from an element and moving diagonally down the triangle, and then making a turn and summing vertically. The specific identity is given by:\[{ }^{r}C_{r} + { }^{r+1}C_{r} + { }^{r+2}C_{r} + \ldots + { }^{r+s}C_{r} = { }^{r+s+1}C_{r+1}\]
Here, these represent binomial coefficients. The sum starts at \(^{r}C_{r}\) and continues to \(^{r+s}C_{r}\), resulting in the sum \(^{r+s+1}C_{r+1}\). It simplifies evaluating series of coefficients efficiently by identifying these patterns. This identity is crucial in problems dealing with sequences of binomial coefficients.
Combinatorial Identities
Combinatorial identities form the core of solving problems in combinatorics and are extensive in areas such as probability, statistics, and algorithms. They are equations involving binomial coefficients and other combinatorial objects and express the equality of two expressions counting the same thing in different ways.
Some typical properties include symmetry and additive identities:
Recognizing these identities enables solving complex problems swiftly by reducing them to manageable components. They play a vital role in sharpening critical thinking and mathematical reasoning.
Some typical properties include symmetry and additive identities:
- **Symmetry Identity**: \({ }^{n}C_{k} = { }^{n}C_{n-k}\), showing that choosing \(k\) elements is the same as excluding \(k\) elements.
- **Addition Identity**: \({ }^{n}C_{k} = { }^{n-1}C_{k} + { }^{n-1}C_{k-1}\), reflects the principle of adding two cases together.
Recognizing these identities enables solving complex problems swiftly by reducing them to manageable components. They play a vital role in sharpening critical thinking and mathematical reasoning.
Binomial Theorem
The Binomial Theorem is a foundational mathematical formula that expands any powers of a binomial expression: \((a + b)^n\). It expresses the result as a sum involving binomial coefficients multiplied by powers of the chosen terms. The formula is:\[(a + b)^n = \sum_{k=0}^{n} { }^{n}C_{k} \, a^{n-k} \, b^{k}\]
This powerful theorem provides a bridge between algebra and combinatorics. It helps simplify expressions and calculate expansions without multiplying everything out explicitly, saving time and minimizing errors. The coefficients \(^{n}C_{k}\) are directly derived from the elements of Pascal's Triangle and encode the idea of selecting elements in different ways from a set.
The Binomial Theorem also lays the groundwork for learning more advanced mathematical concepts, including probability distribution and calculus.
This powerful theorem provides a bridge between algebra and combinatorics. It helps simplify expressions and calculate expansions without multiplying everything out explicitly, saving time and minimizing errors. The coefficients \(^{n}C_{k}\) are directly derived from the elements of Pascal's Triangle and encode the idea of selecting elements in different ways from a set.
The Binomial Theorem also lays the groundwork for learning more advanced mathematical concepts, including probability distribution and calculus.
Other exercises in this chapter
Problem 72
Let \(R=(5 \sqrt{5}+11)^{2 \mathrm{n}+1}\) and \(f=R-[R]\) where [ ] denotes the greatest integer function. Then \(R f=\) (A) \(2^{2 n+1}\) (B) \(\mathrm{W} 2^{
View solution Problem 73
Let \(n\) and \(k\) be positive integers such that \(n \geq \frac{k(k+1)}{2}\) The number of solutions \(\left(x_{1}, x_{2}, \ldots, x_{k}\right), x_{1} \geq 1,
View solution Problem 77
If \((1+x)^{15}=C_{0}+C_{1} x+C_{2} x^{2}+\ldots+C_{15} x^{15}\), then the value of \(C_{2}+2 C_{3}+3 C_{4}+\ldots+14 C_{15}\) is (A) 219923 (B) 16789 (C) 21998
View solution Problem 79
The coefficient of \(x^{30}\) in the expression \((1+x)^{1000}+2 x(1+x)^{999}+3 x^{2}(1+x)^{998}+\ldots+1001 x^{1000}\) is (A) \({ }^{1000} \mathrm{C}_{50}\) (B
View solution