Binomial questions strike fear into the heart of many Mathematics Extension 1 students. These 4 strategies for solving a binomial proof will give you confidence ahead of your exams.

Often binomial proofs can be the most difficult questions in the Maths Extension 1 exam, with students struggling to approach these complex proofs. However, there are certain strategies that you can use to tackle these questions. Your first step is to expand \( (1+x){^n}\), or a similar expression if otherwise stated in the question.

\((1+x){^n}= {{n}\choose{0}} + {{n}\choose{1}}x + {{n}\choose{2}}x{^2}+ \dots + {{n}\choose{k}}x{^k} + \dots {{n}\choose{n}}x{^n}\)

Your next step is to consider the four strategies below.

When to use it: Examine the final term in your expansion and see if replacing it with a number will make your expansion *look* like the answer. However, if you are unsure then it is fine to use trial and error. This won’t take too long as the only substitutions I’ve only ever seen required are x= -1, -1, 0, 1 or 2. Consider the example below.

**Question 1:**

Use the expansion \( (1+x){^n} \) to prove

\({{n}\choose{0}}-2{{n}\choose{2}}+4{{n}\choose{2}}+\dots+(-2){^n}{{n}\choose{n}} \)

has a value of 1 when * n* is even and a value of -1 when

When to use it: Look for signs of differentiation in the answer, most notably anything to the power of (*n*-1), such as *n*2* ^{n-1}*. This indicates that you must differentiate

**Question 2:**

Use the expansion \( (1+x){^n}\) to prove that

\(2{{n}\choose{2}} +3{{n}\choose{3}}+4{{n}\choose{4}}+\dots+k{{n}\choose{k}}+ \dots +n {{n}\choose{n}}=n2{^{n-1}}-n\)

When to use it: Look for signs of integration. Something raised to the power of (* n* + 1) in the answer is a clear sign. When integrating, consider using a definite integral; the limits are often easy numbers such as x = 0, ±1, or ±2. Note that you may be required to integrate twice. In this situation, integrate without limits, which will result in a +

**Question 3:**

Use the expansion \((1+x){^n} \) to prove

\(\frac{1}{2}{{n}\choose{1}}+\frac{1}{3}{{n}\choose{2}}+\frac{1}{4}{{n}\choose{3}}+ \dots+ \frac{1}{k+1}{{n}\choose{k}}+ \dots +\frac{1}{n+1}{{n}\choose{n}}= \frac {2{^{n+1}}-n-2}{n+1} \)

This method involves rewriting a binomial expression in a different way, such as \( (1+x){^{n+2}}=(1+x){^n}(1+x){^2}\) followed by equating coefficients of a specific term, such as *x*^{2.}

When to use this method: First, it is important to understand what \({{n}\choose{r}} \) means in terms of coefficients. \({{n}\choose{r}} \) is the coefficient of \( x{^r}\)*^{ }*in the expansion of \((1=x){^n}\),\({{n+2}\choose{6}}\) is the coefficient of\( x{^6}\)

With this in mind, we can recognize that we must expand the binomial in the question in 2 ways when the power of the expansions differ. For example, if the question is to prove \( {{20}\choose{15}}={{19}\choose{14}}+{{19}\choose{15}}\), then you must consider the expansion \((1+x){^{20}}=(1+x){^{19}}(1+x) \) and then equate coefficients of like terms.

**Question 4: **

Use the expansion of \((1+x){^{n+2}}\) to show

\({{n+2}\choose{r}}={{n-1}\choose{r}}+3{{n-1}\choose{r-1}}+3{{n-1}\choose{r-2}}+{{n-1}\choose{r-3}} \)

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