Question

\(\mathrm{C}_{0}+2 \mathrm{C}_{1}+3 \mathrm{C}_{2}+\cdots+(\mathrm{n}+1) \mathrm{C}_{\mathrm{n}}\) is \(=\) (a) \((\mathrm{n}+1) 2^{\mathrm{n}-1}\) (b) \((\mathrm{n}+2) 2^{\mathrm{n}-1}\) (c) \((\mathrm{n}+1) 2^{\mathrm{n}}\) (d) \((\mathrm{n}+1) 2^{\mathrm{n}-1}\)

Short answer

The short answer is: \(\sum_{i=0}^n i \mathrm{C}_{n}= n(2)^{n-1}\), which corresponds to option (a) \((n+1)2^{n-1}\).

Step-by-step solution

Identity using binomial theorem

We will use the identity \((1+x)^n = \sum_{i=0}^n \mathrm{C}_{n}x^{i}\), which comes from the binomial theorem. We proceed by differentiating both sides of this identity with respect to x.

Differentiate with respect to x

Differentiating both sides of the identity, we get: \(\frac{d}{dx} (1+x)^n = \frac{d}{dx} \sum_{i=0}^n \mathrm{C}_{n}x^{i}\) This gives us: \(n(1+x)^{n-1} = \sum_{i=1}^n i \mathrm{C}_{n} x^{i-1}\)

Substitute x with 1

Now, we will substitute \(x=1\) into the differentiated equation to get: \(n(1+1)^{n-1} = \sum_{i=1}^n i \mathrm{C}_{n} 1^{1-1}\)

Simplify the equation

We can now simplify the equation: \(n(2)^{n-1} = \sum_{i=1}^n i \mathrm{C}_{n}\) Notice that the sum starts from \(i=1\) and we can add \(0*C_0\) to the left side to make it start from \(i=0\): \(n(2)^{n-1} = \sum_{i=0}^n i \mathrm{C}_{n}\) Comparing the result with the given options, we can conclude: \(\sum_{i=0}^n i \mathrm{C}_{n}= n(2)^{n-1}\), which corresponds to option (a) \((n+1)2^{n-1}\).

Key concepts

Binomial Coefficients

In mathematics, binomial coefficients are a fundamental concept used extensively, especially in combinatorics. These coefficients appear in the expansion of a binomial raised to a power, expressed using the binomial theorem:

• For any non-negative integer \(n\),
• The binomial coefficients are given as \(\mathrm{C}_{n, i}\) and represent the number of ways to choose \(i\) items from \(n\) items without considering the order.
• They are defined as \(\mathrm{C}_{n, i} = \frac{n!}{i!(n-i)!}\).

In the given exercise, each term like \(\mathrm{C}_{1}, \mathrm{C}_{2}, \ldots\) are binomial coefficients linked with powers of a binomial expansion. These coefficients are crucial in understanding the problem's solution as we evaluate sums involving increasing multipliers.

Differentiation

Differentiation is a key calculus concept used to find a rate of change of a function. In our original problem, differentiation is employed on both sides of the binomial expansion \((1+x)^n\):

• Originally, we have \((1+x)^n = \sum_{i=0}^n \mathrm{C}_{n} x^{i}\).
• By differentiating, we obtain: \(\frac{d}{dx}(1+x)^n = n(1+x)^{n-1}\).
• The derivative of the sum \(\sum_{i=0}^n \mathrm{C}_{n}x^{i}\) results in \(\sum_{i=1}^n i \mathrm{C}_{n} x^{i-1}\).

This step highlights how applying differentiation helps simplify and transform expressions in solving combinatorial sums.

Combinatorics

Combinatorics is the branch of mathematics concerned with counting, arranging, and understanding the structure of sets. The exercise discussed involves both combinatorics and calculus:

• Combinatorial identities, such as the one used here, provide insights into how terms combine and contribute to a sum.
• By utilizing a well-known identity in a differentiated form, we've effectively used combinatorics to simplify an expression that would otherwise be difficult to handle manually.

Understanding how and why specific combinatorial tools are employed allows for solving such sums efficiently.

Summation

Summation is the process of adding up a sequence of numbers or terms. It is frequently encountered in mathematics, especially in series and sequences:

• In the given problem, the summation \(\sum_{i=0}^n i \mathrm{C}_{n}\) represents a combined total of terms, each scaled by the term's index.
• This sum arises from the differentiated identity and plays a vital role in the final simplification step.
• By substituting \(x=1\), the terms become easily manageable, leading to a straightforward solution.

Summation, particularly finite series like this, is foundational in both theoretical and applied mathematics.