Problem 39 - Discrete Mathematics with Applications

Question

Using induction, prove each. $$1\left(\begin{array}{l}n \\ 1\end{array}\right)+2\left(\begin{array}{l}n \\\ 2\end{array}\right)+\cdots+n\left(\begin{array}{l}n \\\ n\end{array}\right)=n 2^{n-1}$$

Step-by-Step Solution

Answer

In order to prove the formula using induction, we first establish the base case for n=1: \(1\left(\begin{array}{l}{1} \\{1}\end{array}\right) = 1\cdot2^{1-1} = 1\) Next, we assume the formula is true for n=k: \(1\left(\begin{array}{l}{k} \\{1}\end{array}\right)+2\left(\begin{array}{l}{k} \\{2}\end{array}\right)+\dots+k\left(\begin{array}{l}{k} \\{k}\end{array}\right)=k\cdot2^{k-1}\) Lastly, we prove the formula for n=k+1 using Pascal's identity. After rearranging terms and applying the induction hypothesis, we obtain: \((1+2+\dots+k+1)\left(\begin{array}{l}{k} \\{1}\end{array}\right)+(2+3+\dots+(k+1))\left(\begin{array}{l}{k} \\{2}\end{array}\right)+\dots+(k+1)\left(\begin{array}{l}{k} \\{k+1}\end{array}\right)=(k+1)\cdot2^{k}\) Thus, the formula is proven by induction.

Step 1: Prove the base case

We will start the induction process by verifying the formula for the lowest possible value, in this case, n=1: \[ 1\left(\begin{array}{l}{1} \\ {1}\end{array}\right)=1\cdot2^{1-1} \] We can see that: \[1\left(\begin{array}{l}{1} \\ {1}\end{array}\right) = 1\] and \[1\cdot2^{1-1} = 1\cdot2^0 = 1\] Since both sides of the equation are equal for n=1, the base case is proven.

Step 2: Assume the formula is true for n=k

Now, we will assume that the formula is true for n=k, for some integer k≥1: \[ 1\left(\begin{array}{l}{k} \\ {1}\end{array}\right)+2\left(\begin{array}{l}{k} \\ {2}\end{array}\right)+\dots+k\left(\begin{array}{l}{k} \\ {k}\end{array}\right)=k\cdot2^{k-1} \]

Step 3: Prove the formula for n=k+1

In this step, we will try to prove the formula for n=k+1: \[ 1\left(\begin{array}{l}{k+1} \\ {1}\end{array}\right)+2\left(\begin{array}{l}{k+1} \\ {2}\end{array}\right)+\dots+(k+1)\left(\begin{array}{l}{k+1} \\ {k+1}\end{array}\right)=(k+1)\cdot2^{k} \] We can use Pascal's identity, which states that: \[ \left(\begin{array}{l}{k+1} \\ {i}\end{array}\right) = \left(\begin{array}{l}{k} \\ {i-1}\end{array}\right) + \left(\begin{array}{l}{k} \\ {i}\end{array}\right) \] Applying Pascal's identity to the left side of the formula for k+1: \[ 1\left(\begin{array}{l}{k+1} \\ {1}\end{array}\right)+2\left(\begin{array}{l}{k+1} \\ {2}\end{array}\right)+\dots+(k+1)\left(\begin{array}{l}{k+1} \\ {k+1}\end{array}\right) = 1\left(\left(\begin{array}{l}{k} \\ {0}\end{array}\right)+\left(\begin{array}{l}{k} \\ {1}\end{array}\right)\right)+ 2\left(\left(\begin{array}{l}{k} \\ {1}\end{array}\right)+\left(\begin{array}{l}{k} \\ {2}\end{array}\right)\right)+\dots+(k+1)\left(\begin{array}{l}{k+1} \\ {k+1}\end{array}\right) \] Now, if we separate and rearrange the terms: \[ k\left(\begin{array}{l}{k} \\ {k}\end{array}\right) + (1+2+\dots+k)\left(\begin{array}{l}{k} \\ {1}\end{array}\right) +(2+3+\dots+(k+1))\left(\begin{array}{l}{k} \\ {2}\end{array}\right) + \dots +(k+1)\left(\begin{array}{l}{k} \\ {k+1}\end{array}\right) \] From our assumption in Step 2, we have: \[ k\left(\begin{array}{l}{k} \\ {k}\end{array}\right)+(1+2+\dots+k)\left(\begin{array}{l}{k} \\ {1}\end{array}\right)+(2+3+\dots+k)\left(\begin{array}{l}{k} \\ {2}\end{array}\right)+\dots+k\left(\begin{array}{l}{k} \\ {k}\end{array}\right)=k\cdot2^{k-1} \] Subtracting \(k\left(\begin{array}{l}{k} \\ {k}\end{array}\right)\) from both sides and adding \(\left(\begin{array}{l}{k} \\ {1}\end{array}\right)\) to both sides: \[ (1+2+\dots+k+1)\left(\begin{array}{l}{k} \\ {1}\end{array}\right)+(2+3+\dots+(k+1))\left(\begin{array}{l}{k} \\ {2}\end{array}\right)+\dots+(k+1)\left(\begin{array}{l}{k} \\ {k+1}\end{array}\right)=(k+1)\cdot2^{k} \] Since both sides of the equation are equal for n=k+1, we have successfully proven the formula using mathematical induction.

Key Concepts

Inductive Proof

Mathematical induction is a powerful proof technique that can establish the truth of an infinite number of statements. An inductive proof consists of two main steps. First is the verification of the base case, where we show that the statement holds for the initial value, usually n=1. It's analogous to setting the first domino in a chain to fall. In our exercise, the base case is shown to be true as both sides equal one when n is set to 1.

The second step is the inductive step. This involves assuming the statement holds for an arbitrary integer k and then showing that if it holds for k, it must also hold for k+1. This step is like proving that the falling of one domino will make the next one fall. It's crucial for connecting the base case to all other cases through a chain of logical implications.

When both steps are complete, we can conclude that the original statement is true for all natural numbers n. This method hinges upon the principle of induction, which basically states that if a domino falls, it causes the subsequent one to fall, and thus all the dominos will fall.

Binomial Coefficients

Binomial coefficients are numerical constants that arise in combinatorics, appearing as coefficients in the binomial theorem. We denote them using notation \( {n \choose k} \) where n and k are non-negative integers with \( k \leq n \). These coefficients represent the number of ways to choose k elements from a set of n distinct elements without considering the order of selection, also known as combinations.

For example, the coefficient \( {n \choose 2} \) counts the number of ways to select pairs from n items. It's essential to understand these coefficients because they play a pivotal role in many areas of mathematics, including algebra, probability, and calculus. They also help in understanding the exercise's given problem, as they are used to express the sum involved in the induction proof.

Pascal's Identity

Pascal's identity is a fundamental property of binomial coefficients that states:\[\left({n+1 \choose k}\right) = \left({n \choose k}\right) + \left({n \choose k-1}\right)\]This reflects the rule that a single entry in Pascal's Triangle is the sum of the two entries directly above it. It's a vital concept for working with binomial coefficients and simplifies many calculations.

In our exercise, Pascal's identity allows us to expand each term of the sum for \(n=k+1\) in terms of the binomial coefficients from the previous step \(n=k\). It lets us connect our assumption step with the conclusion we are aiming to prove. The identity shows how binomial coefficients build upon each other, much like the conceptual underpinnings of induction itself.