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Chapter 5: Problem 1

Compute the sum $$ \sum_{k=0}^{3 n-1}(-1)^{k}\left(\begin{array}{c} 6 n \\ 2 k+1 \end{array}\right) 3^{k} $$

Short Answer

Expert verified
Question: Compute and simplify the result of the given finite series: $$\sum_{k=0}^{3n - 1} (-1)^k \binom{6n}{2k + 1} 3^k$$ Follow the steps outlined in the solution above, applying the properties of binomial coefficients and the geometric series sum. Write your final simplified answer.

Step by step solution

01

Write the sum in a more readable form

The given sum can be written as: $$ \sum_{k=0}^{3n - 1} (-1)^k \binom{6n}{2k + 1} 3^k $$
02

Split the sum into two sub-sums to better identify a pattern

Let's split the sum into two sub-sums based on even and odd terms: $$ \sum_{k=0}^{3n - 1} (-1)^k \binom{6n}{2k + 1} 3^k = \sum_{k=0 \text{; even}}^{3n - 1} (-1)^k \binom{6n}{2k + 1} 3^k - \sum_{k=0 \text{; odd}}^{3n - 1} (-1)^k \binom{6n}{2k + 1} 3^k $$
03

Factor out \(3^k\) from each term in the sub-sums

We can factor out \(3^k\) from each term in the two sub-sums as follows: $$ = 3^0 \left( \sum_{k=0 \text{; even}}^{3n - 1} (-1)^k \binom{6n}{2k + 1} \right) - 3^1 \left( \sum_{k=0 \text{; odd}}^{3n - 1} (-1)^k \binom{6n}{2k + 1} \right) $$
04

Replace \((-1)^k\) with \(\binom{-1}{k}\) using the binomial theorem

Since \((-1)^k = \binom{-1}{k}\), we can replace in our sum: $$ = 3^0 \left( \sum_{k=0 \text{; even}}^{3n - 1} \binom{-1}{k} \binom{6n}{2k + 1} \right) - 3^1 \left( \sum_{k=0 \text{; odd}}^{3n - 1} \binom{-1}{k} \binom{6n}{2k + 1} \right) $$
05

Use the identity of Vandermonde's Convolution for the binomial coefficients

Using the identity for Vandermonde's Convolution: $$ \binom{m+n}{r} = \sum_{k=0}^{r} \binom{m}{k} \binom{n}{r-k} $$ Our sum can be written as: $$ = \binom{6n}{1} - 3 \binom{6n-1}{0} $$ Now let's compute the binomial coefficients and simplify the result.
06

Compute the binomial coefficients and simplify the result

$$ \binom{6n}{1} = \frac{6n!}{1!(6n-1)!} = 6n $$ and $$ \binom{6n-1}{0} = \frac{(6n-1)!}{0!(6n-1)!} = 1 $$ So the final result is: $$ = 6n - 3 \cdot 1 = 6n - 3 $$

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Key Concepts

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

Binomial Coefficients
Binomial coefficients are a central concept in combinatorics and appear frequently in algebra and calculus. In simplest terms, a binomial coefficient \(\binom{n}{k}\) tells us the number of ways to choose a subset of \(k\) elements from a larger set of \(n\) distinct elements, without considering the order of selection. This value is often referred to as \

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Most popular questions from this chapter

Sum the following two \(n\) -term series for \(\theta=30^{\circ}\) : (i) \(1+\frac{\cos \theta}{\cos \theta}+\frac{\cos (2 \theta)}{\cos ^{2} \theta}+\frac{\cos (3 \theta)}{\cos ^{3} \theta}+\cdots+\frac{\cos ((n-1) \theta)}{\cos ^{n-1} \theta}\), and (ii) \(\cos \theta \cos \theta+\cos ^{2} \theta \cos (2 \theta)+\cos ^{3} \theta \cos (3 \theta)+\cdots+\cos ^{n} \theta \cos (n \theta)\). (Crux Mathematicorum, 2003)

Let \(n \geq 3\) be an integer and \(z=\cos \frac{2 \pi}{n}+i \sin \frac{2 \pi}{n} .\) Consider the sets $$ A=\left\\{1, z, z^{2}, \ldots, z^{n-1}\right\\} $$ and $$ B=\left\\{1,1+z, 1+z+z^{2}, \ldots, 1+z+\ldots+z^{n-1}\right\\} . $$ Determine \(A \cap B\).

Let \(n>2\) be an integer and \(f: \mathbb{R}^{2} \rightarrow \mathbb{R}\) a function such that for every regular \(n\) -gon \(A_{1} A_{2} \cdots A_{n}\), $$ f\left(A_{1}\right)+f\left(A_{2}\right)+\cdots+f\left(A_{n}\right)=0 $$ Prove that \(f\) is identically zero. (Romanian Mathematical Olympiad-Final Round, 1996 )

The points \(A_{1}, A_{2}, \ldots, A_{10}\) are equally distributed on a circle of radius \(R\) (in that order). Prove that \(A_{1} A_{4}-A_{1} A_{2}=R\).

Given a positive integer \(n\), it can be shown that every complex number of the form \(r+\) si, where \(r\) and \(s\) are integers, can be uniquely expressed in the base \(-n+i\) using the integers \(1,2, \ldots, n^{2}\) as digits. That is, the equation $$ r+s i=a_{m}(-n+i)^{m}+a_{m-1}(-n+i)^{m-1}+\ldots+a_{1}(-n+i)+a_{0} $$ is valid for a unique choice of nonnegative integer \(m\) and digits \(a_{0}, a_{1}, \ldots, a_{m}\) chosen from the set \(\left\\{0,1,2, \ldots, n^{2}\right\\}\), with \(a_{m} \neq 0 . \mathrm{We}\) write $$ r+s i=\left(a_{m} a_{m-1} \ldots a_{1} a_{0}\right)_{-n+i} $$ to denote the base- \((-n+i)\) expansion of \(r+\) si. There are only finitely many integers \(k+0 i\) that have four-digit expansions $$ k=\left(a_{3} a_{2} a_{1} a_{0}\right)_{-3+i}, a_{3} \neq 0 $$ Find the sum of all such \(k\).
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