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(For students who have studied calculus) a. Explain how the equation below follows from the binomial theorem: $$ (1+x)^{n}=\sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right) x^{k} $$ b. Write the formula obtained by taking the derivative of both sides of the equation in part (a) with respect to \(x\). c. Use the result of part (b) to derive the formulas below. (i) \(2^{n-1}=\frac{1}{n}\left[\left(\begin{array}{l}n \\\ 1\end{array}\right)+2\left(\begin{array}{l}n \\\ 2\end{array}\right)+3\left(\begin{array}{l}n \\\ 3\end{array}\right)+\cdots+n\left(\begin{array}{l}n \\\ n\end{array}\right)\right]\) (ii) \(\sum_{k=1}^{n} k\left(\begin{array}{l}n \\\ k\end{array}\right)(-1)^{k}=0\) d. Express \(\sum_{k=1}^{n} k\left(\begin{array}{l}n \\ k\end{array}\right) 3^{k}\) in closed form (without using a summation sign or \(\cdots\) ).

Step by step solution

01

Review the binomial theorem

The binomial theorem states that for any positive integer n and any real numbers x and y: \((x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{k} y^{n - k}\) In this case, we have \((1+x)^n\), so we can rewrite the binomial theorem formula as: \((1+x)^n = \sum_{k=0}^{n} \binom{n}{k} x^{k} (1)^{n-k}\) Since \(1^{n-k}\) = 1, we get: \((1+x)^n = \sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right) x^{k}\) #b. Derivative of both sides of the equation#

02

Take the derivative

We now take the derivative of both sides of the equation with respect to x: \(\frac{d}{dx} (1+x)^n = \frac{d}{dx} \sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right) x^{k}\) Applying the power rule on the left side and differentiating term by term on the right side, we get: \(n(1+x)^{n-1} = \sum_{k=1}^{n} k\left(\begin{array}{l} n \\ k \end{array}\right) x^{k-1}\) #c. Derive the given formulas# #(i) Derive the first formula#

03

Plug in x = 1

To derive the first formula, substitute x = 1 into the result of part (b): \(n(1+1)^{n-1} = \sum_{k=1}^{n} k\left(\begin{array}{l} n \\ k \end{array}\right) (1)^{k-1}\) Simplify the left side and replace (1)^{k-1} with 1: \(n(2)^{n-1} = \sum_{k=1}^{n} k\left(\begin{array}{l} n \\ k \end{array}\right)\) Factor out the \(\frac{1}{n}\) on both sides to get the desired formula: \(2^{n-1} = \frac{1}{n}\left(\sum_{k=1}^{n} k\left(\begin{array}{l} n \\ k \end{array}\right)\right)\) #(ii) Derive the second formula#

04

Plug in x = -1

To derive the second formula, substitute x = -1 into the result of part (b): \(n(1-1)^{n-1} = \sum_{k=1}^{n} k\left(\begin{array}{l} n \\ k \end{array}\right) (-1)^{k-1}\) Since (1-1)^{n-1} = 0, the equation becomes: \(0 = \sum_{k=1}^{n} k\left(\begin{array}{l} n \\ k \end{array}\right) (-1)^{k}\) This is the second formula we wanted to derive. #d. Express the sum in closed form#

05

Plug in x = 3

To find the closed form for the given sum, substitute x = 3 into the result of part (b): \(n(1+3)^{n-1} = \sum_{k=1}^{n} k\left(\begin{array}{l} n \\ k \end{array}\right) (3)^{k-1}\) Simplify the left side: \(n(4)^{n-1} = \sum_{k=1}^{n} k\left(\begin{array}{l} n \\ k \end{array}\right) 3^{k-1}\) Now, we have expressed the sum as: \(\sum_{k=1}^{n} k\left(\begin{array}{l} n \\ k \end{array}\right) 3^{k} = n(4)^{n-1}\)

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Key Concepts

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

Calculus

Calculus is a branch of mathematics that helps us understand changes between values. It has two main parts: differentiation and integration.
Differentiation deals with how things change and is all about rates of change, while integration focuses on accumulation. In this exercise, we use differentiation when taking the derivative of a function.
This is why we see how the expression \((1+x)^n\) can change by applying the rules of derivative calculus.

Derivative

A derivative refers to the rate at which a function changes at any given point. It's like finding how fast a car is going when you look at the speedometer.
In math, when you want to find the derivative of \((1+x)^n\) with respect to \(x\), you apply the power rule of derivatives:

• If you have a function \(x^k\), its derivative is \(kx^{k-1}\).

This gives us\(n(1+x)^{n-1}\) on the left side. On the right, you differentiate each term of the sum one by one, which leads to \(\sum_{k=1}^{n} k\left(\begin{array}{l} n \ k \end{array}\right) x^{k-1}\).
This way, derivatives help us break down the function into simpler parts, keeping everything systematic and organized.

Closed form expression

A closed form expression provides a neat and concise way to write something that would otherwise be long and complex. It's like a shortcut for big calculations.
This exercise explores how to transform a sum into a closed form expression. For example, instead of using the lengthy summation, we use the result of derivative calculation to write

• \(\sum_{k=1}^{n} k\left(\begin{array}{l} n \ k \end{array}\right) 3^k = n(4)^{n-1}\).

This eliminates the need for the summation symbol, making it easier to understand and work with. Simplifying expressions through closed forms can make problem-solving much quicker.

Summation formula

A summation formula is used to express the sum of a sequence. It's like adding up all the terms in a series without having to write them all out individually.
In our case, we start with the binomial theorem expression:

• \(\sum_{k=0}^{n}\left(\begin{array}{l} n \ k \end{array}\right) x^k\)

After deriving and substituting values, we can find specific summation formulas such as:

• \(\sum_{k=1}^{n} k\left(\begin{array}{l} n \ k \end{array}\right)(-1)^k = 0\).

Summation formulas help to systematically handle large series, breaking them down into understandable parts without manually adding every term.