Problem 34 - Discrete Mathematics with Applications

Question

Evaluate each sum. \(1\left(\begin{array}{l}n \\ 1\end{array}\right)+2\left(\begin{array}{l}n \\\ 2\end{array}\right)+3\left(\begin{array}{l}n \\\ 3\end{array}\right)+\cdots+n\left(\begin{array}{l}n \\ n\end{array}\right)\) (Hint: Let \(S\) denote the sum. Use \(S\) and the sum in the reverse order to compute \(2 S .\))

Step-by-Step Solution

Answer

The short answer to the given sum is: $$ S = \frac{n(2^n-1)}{2} $$

Step 1: Expression of the given sum (\(S\))

First of all, let's write the given sum \(S\) as an expression. The sum is given as: $$ S = \sum_{k=1}^n k\binom{n}{k} $$

Step 2: Reverse order sum

Now, let's write the reverse order sum (notation: \(S_{rev}\)) by changing the order of the summands. $$ S_{rev} = n\binom{n}{n-1}+(n-1)\binom{n}{n-2}+(n-2)\binom{n}{n-3}+ ...+ 2\binom{n}{1}+1\binom{n}{0} $$

Step 3: Relation between binomial coefficients

Now, we will use the following relation for binomial coefficients: \[ \binom{n}{k} = \binom{n}{n-k} \] Applying this property on \(S_{rev}\), we get: $$ S_{rev} = n\binom{n}{1} + (n-1)\binom{n}{2} + (n-2)\binom{n}{3}+ ...+ 2\binom{n}{n-1}+1\binom{n}{n} $$ It's clear that \(S_{rev}\) has the same terms as \(S\), but just in the reverse order.

Step 4: Sum \(S\) and reverse sum \(S_{rev}\)

Now, we will compute the sum (\(S + S_{rev}\)) to find a simple relation for the sum \(S\). $$ S + S_{rev} = \sum_{k=1}^n k\binom{n}{k} + \sum_{k=1}^n (n-k)\binom{n}{k} $$ Which also can be rewritten as: $$ S + S_{rev} = \sum_{k=1}^n (k+(n-k))\binom{n}{k} = n\sum_{k=1}^n \binom{n}{k} $$

Step 5: Using binomial theorem

We can use the binomial theorem to find the sum of all binomial coefficients: $$ \sum_{k=0}^n \binom{n}{k} = 2^n $$ Since we already have \(\binom{n}{0}\) in the sum above, let's substract the term from it and rewrite our sum with \(k\) from 1 to \(n\): $$ \sum_{k=1}^n \binom{n}{k} = 2^n - 1 $$

Step 6: Finding the sum S

Now, we can use this result to find the value of the sum \(S\). From our previous conclusion: $$ S + S_{rev} = n(2^n - 1) $$ But since \(S\) and \(S_{rev}\) share the same terms, they must be equal: $$ S = S_{rev} $$ Thus, $$ 2S = n(2^n - 1) $$ Finally, we can evaluate the sum \(S\) as: $$ S = \frac{n(2^n-1)}{2} $$ This is the result of the asked sum.

Key Concepts

Understanding Binomial Coefficients

Binomial coefficients are an essential concept in combinatorics, representing coefficients in the binomial theorem expansion. When you expand \((x + y)^n\), binomial coefficients are the numbers \(\binom{n}{k}\), where \(n\) is a non-negative integer and \(k\) is the index of the term in the expansion. These coefficients are crucial in determining how many ways to pick \(k\) objects from a set of \(n\) objects, without regard to the order.

Key properties include:

• Symmetry: \(\binom{n}{k} = \binom{n}{n-k}\).
• Pascal's Rule: \(\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}\).

Binomial coefficients are often visualized in a triangular arrangement called Pascal's Triangle, where each number is the sum of the two numbers directly above it.

Calculating the Sum of Binomial Coefficients

The sum of binomial coefficients is a powerful tool in algebra and combinatorics. When considering the expression for the sum \(S = \sum_{k=1}^n k\binom{n}{k}\), it may seem complex at first.However, by flipping the order as done in the reverse sum allows for cancellation and simplification.

Another crucial result from the binomial theorem tells us:\[\sum_{k=0}^n \binom{n}{k} = 2^n\]By excluding the term \(\binom{n}{0}\), we find that:\[\sum_{k=1}^n \binom{n}{k} = 2^n - 1\]This manipulation aids in simplifying problems requiring the sum of certain types of binomial coefficient expressions.

This approach reflects an instance where understanding basic principles from the binomial theorem aids in more complex calculations.

Exploring Combinatorial Identities

Combinatorial identities are equations that hold true for all natural numbers and involve binomial coefficients. They provide powerful tools to solve problems involving combinations and arrangements. These identities stem from the properties of binomial coefficients and their applications extend beyond the binomial theorem.

Some common combinatorial identities include:

• Vandermonde's Identity: \(\binom{m+n}{r} = \sum_{k=0}^r \binom{m}{k}\binom{n}{r-k}\)
• Binomial Theorem: \((x+y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k} y^k\)

Identities like these allow us to transform and simplify expressions, made even more powerful when combined with the symmetry and additive properties of binomial coefficients. They form a backbone for deeper exploration into permutations, combinations, and advanced algebraic structures.