Which of the following correctly proves Pascal's rule nr+nr+1=n+1r+1?

Write the two coefficients with factorials: nr+nr+1=(n!)/(r!(n-r)!)+(n!)/((r+1)!(n-r-1)!) Use a common denominator: =(n! (r+1)+n! (n-r))/((r+1)!(n-r)!) Recall the binomial theorem: (a+b)^n=_r=0^nnra^n-rb^r Recall the factorial form of the coefficient: nr=(n!)/(r! (n-r)!) Remember the symmetry of the coefficients: nr=nn-r Check the number of terms: (a+bx)^n has n+1 terms Watch the powers add up to n: power of a + power of bx = n Recall Pascal's triangle relation: nr=n-1r-1+n-1r Note the end coefficients: n0=1, nn=1 Be careful with the signs: (-1)^r>0 when r is even, <0 when r is odd Recall the binomial theorem: (a+b)^n=_r=0^nnra^n-rb^r Recall the factorial form of the coefficient: nr=(n!)/(r! (n-r)!) Remember the symmetry of the coefficients: nr=nn-r Check the number of terms: (a+bx)^n has n+1 terms Simplify the numerator: =(n! (n+1))/((r+1)!(n-r)!)=((n+1)!)/((r+1)!(n-r)!)=n+1r+1 Answer: Add over a common denominator to get (n+1)! on top, i.e. n+1r+1.

Which of the following correctly proves that _r=0^nnr=2^n?

Start from the binomial theorem: (1+x)^n=_r=0^nnrx^r Substitute x=1: (1+1)^n=_r=0^nnr(1)^r Recall the binomial theorem: (a+b)^n=_r=0^nnra^n-rb^r Recall the factorial form of the coefficient: nr=(n!)/(r! (n-r)!) Remember the symmetry of the coefficients: nr=nn-r Check the number of terms: (a+bx)^n has n+1 terms Watch the powers add up to n: power of a + power of bx = n Recall Pascal's triangle relation: nr=n-1r-1+n-1r Note the end coefficients: n0=1, nn=1 Be careful with the signs: (-1)^r>0 when r is even, <0 when r is odd Recall the binomial theorem: (a+b)^n=_r=0^nnra^n-rb^r Recall the factorial form of the coefficient: nr=(n!)/(r! (n-r)!) Remember the symmetry of the coefficients: nr=nn-r Check the number of terms: (a+bx)^n has n+1 terms Simplify the left side: 2^n=_r=0^nnr Answer: Substitute x=1 into (1+x)^n to obtain 2^n=nr.

Find the value of _r=0^55r2^r.

Recognise the binomial pattern: _r=0^55r2^r=_r=0^55r1^5-r2^r Compare with the binomial theorem: (a+b)^5=_r=0^55ra^5-rb^r Rewrite as a single power: _r=0^55r2^r=(1+2)^5 Recall the binomial theorem: (a+b)^n=_r=0^nnra^n-rb^r Recall the factorial form of the coefficient: nr=(n!)/(r! (n-r)!) Remember the symmetry of the coefficients: nr=nn-r Check the number of terms: (a+bx)^n has n+1 terms Watch the powers add up to n: power of a + power of bx = n Recall Pascal's triangle relation: nr=n-1r-1+n-1r Note the end coefficients: n0=1, nn=1 Be careful with the signs: (-1)^r>0 when r is even, <0 when r is odd Recall the binomial theorem: (a+b)^n=_r=0^nnra^n-rb^r Recall the factorial form of the coefficient: nr=(n!)/(r! (n-r)!) Remember the symmetry of the coefficients: nr=nn-r Evaluate: 3^5=243 Answer: 243

Hard A-Level Binomial expansion Questions

Challenging, exam-style A-Level Binomial expansion questions with worked solutions. Stretch yourself on the hardest coefficient, (1+bx)^n, (a+bx)^n, negative sign problems.

coefficient(1+bx)^n(a+bx)^nnegative signfractional coefficientreordered bracket

A-Level34 questionsStep-by-step solutions

Question 1

8 markschallenging

Explain why the expansion of

(a+bx)^n

(for a positive integer

n

) has exactly

n+1

terms.

Show worked solution

Worked solution

Recall the general term
$\binom{n}{r}a^{n-r}b^r,\quad r=0,1,\dots,n$
Each term of (a+b)^n is set by a value of r. So the possible values of r decide how many terms there are.
Recall the binomial theorem
$(a+b)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r$
This single formula generates every term of the expansion. Getting comfortable writing it down first stops us from missing terms later.
Recall the factorial form of the coefficient
$\binom{n}{r}=\frac{n!}{r!\,(n-r)!}$
The binomial coefficient just counts how many ways r objects can be chosen from n. On your calculator this is the nCr button.
Remember the symmetry of the coefficients
$\binom{n}{r}=\binom{n}{n-r}$
Pascal's triangle is symmetric, so coefficients read the same forwards and backwards. This is a handy check that your numbers are right.
Check the number of terms
$(a+bx)^n \text{ has } n+1 \text{ terms}$
An expansion of power n always has one more term than the power. Counting the terms is a quick way to spot if you have dropped one.
Watch the powers add up to n
$\text{power of }a + \text{power of }bx = n$
In every single term the two powers must add to n. If they don't, an arithmetic slip has crept in.
Recall Pascal's triangle relation
$\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$
Each entry in Pascal's triangle is the sum of the two above it. This is where the coefficients come from without any factorials.
Note the end coefficients
$\binom{n}{0}=1,\qquad \binom{n}{n}=1$
The very first and very last coefficients are always 1. That gives you two terms of the answer for free.
Be careful with the signs
$(-1)^r>0 \text{ when } r \text{ is even, } <0 \text{ when } r \text{ is odd}$
When the bracket contains a minus sign the terms alternate in sign. Track this carefully so no negative is lost.
Recall the binomial theorem
$(a+b)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r$
This single formula generates every term of the expansion. Getting comfortable writing it down first stops us from missing terms later.
Recall the factorial form of the coefficient
$\binom{n}{r}=\frac{n!}{r!\,(n-r)!}$
The binomial coefficient just counts how many ways r objects can be chosen from n. On your calculator this is the nCr button.
Remember the symmetry of the coefficients
$\binom{n}{r}=\binom{n}{n-r}$
Pascal's triangle is symmetric, so coefficients read the same forwards and backwards. This is a handy check that your numbers are right.
Check the number of terms
$(a+bx)^n \text{ has } n+1 \text{ terms}$
An expansion of power n always has one more term than the power. Counting the terms is a quick way to spot if you have dropped one.
Watch the powers add up to n
$\text{power of }a + \text{power of }bx = n$
In every single term the two powers must add to n. If they don't, an arithmetic slip has crept in.
Count the values of r
$r\in\{0,1,2,\dots,n\}\Rightarrow n+1 \text{ values}$
The counter r runs from 0 up to n inclusive. Counting these values, there are n+1 of them, so there are n+1 terms.

Answer

Because r takes the n+1 values 0,1,...,n.

Question 2

9 markschallenging

Which of the following correctly proves Pascal's rule

\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}

Show worked solution

Worked solution

Write the two coefficients with factorials
$\binom{n}{r}+\binom{n}{r+1}=\frac{n!}{r!(n-r)!}+\frac{n!}{(r+1)!(n-r-1)!}$
We start from the factorial definition of each coefficient. The plan is to add the two fractions.
Use a common denominator
$=\frac{n!\,(r+1)+n!\,(n-r)}{(r+1)!(n-r)!}$
We make the denominators the same by adjusting factorials. Then we can add the numerators.
Recall the binomial theorem
$(a+b)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r$
This single formula generates every term of the expansion. Getting comfortable writing it down first stops us from missing terms later.
Recall the factorial form of the coefficient
$\binom{n}{r}=\frac{n!}{r!\,(n-r)!}$
The binomial coefficient just counts how many ways r objects can be chosen from n. On your calculator this is the nCr button.
Remember the symmetry of the coefficients
$\binom{n}{r}=\binom{n}{n-r}$
Pascal's triangle is symmetric, so coefficients read the same forwards and backwards. This is a handy check that your numbers are right.
Check the number of terms
$(a+bx)^n \text{ has } n+1 \text{ terms}$
An expansion of power n always has one more term than the power. Counting the terms is a quick way to spot if you have dropped one.
Watch the powers add up to n
$\text{power of }a + \text{power of }bx = n$
In every single term the two powers must add to n. If they don't, an arithmetic slip has crept in.
Recall Pascal's triangle relation
$\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$
Each entry in Pascal's triangle is the sum of the two above it. This is where the coefficients come from without any factorials.
Note the end coefficients
$\binom{n}{0}=1,\qquad \binom{n}{n}=1$
The very first and very last coefficients are always 1. That gives you two terms of the answer for free.
Be careful with the signs
$(-1)^r>0 \text{ when } r \text{ is even, } <0 \text{ when } r \text{ is odd}$
When the bracket contains a minus sign the terms alternate in sign. Track this carefully so no negative is lost.
Recall the binomial theorem
$(a+b)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r$
This single formula generates every term of the expansion. Getting comfortable writing it down first stops us from missing terms later.
Recall the factorial form of the coefficient
$\binom{n}{r}=\frac{n!}{r!\,(n-r)!}$
The binomial coefficient just counts how many ways r objects can be chosen from n. On your calculator this is the nCr button.
Remember the symmetry of the coefficients
$\binom{n}{r}=\binom{n}{n-r}$
Pascal's triangle is symmetric, so coefficients read the same forwards and backwards. This is a handy check that your numbers are right.
Check the number of terms
$(a+bx)^n \text{ has } n+1 \text{ terms}$
An expansion of power n always has one more term than the power. Counting the terms is a quick way to spot if you have dropped one.
Simplify the numerator
$=\frac{n!\,(n+1)}{(r+1)!(n-r)!}=\frac{(n+1)!}{(r+1)!(n-r)!}=\binom{n+1}{r+1}$
The numerator (r+1)+(n-r)=n+1, so n!(n+1)=(n+1)!. This is exactly \binom{n+1}{r+1}, proving Pascal's rule.

Answer

Add over a common denominator to get (n+1)! on top, i.e. \binom{n+1}{r+1}.

Question 3

8 markschallenging

Which of the following correctly proves that

\displaystyle\sum_{r=0}^{n}\binom{n}{r}=2^n

Show worked solution

Worked solution

Start from the binomial theorem
$(1+x)^n=\sum_{r=0}^{n}\binom{n}{r}x^r$
The sum of all binomial coefficients is what we get by adding the coefficients of (1+x)^n. Substituting a value for x will add them for us.
Substitute x=1
$(1+1)^n=\sum_{r=0}^{n}\binom{n}{r}(1)^r$
Putting x=1 makes every power of x equal to 1, so the right side becomes the plain sum of the coefficients.
Recall the binomial theorem
$(a+b)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r$
This single formula generates every term of the expansion. Getting comfortable writing it down first stops us from missing terms later.
Recall the factorial form of the coefficient
$\binom{n}{r}=\frac{n!}{r!\,(n-r)!}$
The binomial coefficient just counts how many ways r objects can be chosen from n. On your calculator this is the nCr button.
Remember the symmetry of the coefficients
$\binom{n}{r}=\binom{n}{n-r}$
Pascal's triangle is symmetric, so coefficients read the same forwards and backwards. This is a handy check that your numbers are right.
Check the number of terms
$(a+bx)^n \text{ has } n+1 \text{ terms}$
An expansion of power n always has one more term than the power. Counting the terms is a quick way to spot if you have dropped one.
Watch the powers add up to n
$\text{power of }a + \text{power of }bx = n$
In every single term the two powers must add to n. If they don't, an arithmetic slip has crept in.
Recall Pascal's triangle relation
$\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$
Each entry in Pascal's triangle is the sum of the two above it. This is where the coefficients come from without any factorials.
Note the end coefficients
$\binom{n}{0}=1,\qquad \binom{n}{n}=1$
The very first and very last coefficients are always 1. That gives you two terms of the answer for free.
Be careful with the signs
$(-1)^r>0 \text{ when } r \text{ is even, } <0 \text{ when } r \text{ is odd}$
When the bracket contains a minus sign the terms alternate in sign. Track this carefully so no negative is lost.
Recall the binomial theorem
$(a+b)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r$
This single formula generates every term of the expansion. Getting comfortable writing it down first stops us from missing terms later.
Recall the factorial form of the coefficient
$\binom{n}{r}=\frac{n!}{r!\,(n-r)!}$
The binomial coefficient just counts how many ways r objects can be chosen from n. On your calculator this is the nCr button.
Remember the symmetry of the coefficients
$\binom{n}{r}=\binom{n}{n-r}$
Pascal's triangle is symmetric, so coefficients read the same forwards and backwards. This is a handy check that your numbers are right.
Check the number of terms
$(a+bx)^n \text{ has } n+1 \text{ terms}$
An expansion of power n always has one more term than the power. Counting the terms is a quick way to spot if you have dropped one.
Simplify the left side
$2^n=\sum_{r=0}^{n}\binom{n}{r}$
The left side is (1+1)^n=2^n. This proves the sum of the coefficients in row n of Pascal's triangle is 2^n.

Answer

Substitute x=1 into (1+x)^n to obtain 2^n=\sum\binom{n}{r}.

Question 4

8 markschallenging

Find the value of

\displaystyle\sum_{r=0}^{5}\binom{5}{r}2^r

Show worked solution

Worked solution

Recognise the binomial pattern
$\sum_{r=0}^{5}\binom{5}{r}2^r=\sum_{r=0}^{5}\binom{5}{r}1^{5-r}2^r$
The sum looks complicated, but the \binom{5}{r} and the power of 2 are exactly the pieces of a binomial expansion. We write 1^{5-r} to make the pattern clear.
Compare with the binomial theorem
$(a+b)^5=\sum_{r=0}^{5}\binom{5}{r}a^{5-r}b^r$
Matching term by term, we see a=1 and b=2. The whole sum is therefore a single power.
Rewrite as a single power
$\sum_{r=0}^{5}\binom{5}{r}2^r=(1+2)^5$
The sum collapses to (1+2)^5. This is the clever insight that avoids adding six separate terms.
Recall the binomial theorem
$(a+b)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r$
This single formula generates every term of the expansion. Getting comfortable writing it down first stops us from missing terms later.
Recall the factorial form of the coefficient
$\binom{n}{r}=\frac{n!}{r!\,(n-r)!}$
The binomial coefficient just counts how many ways r objects can be chosen from n. On your calculator this is the nCr button.
Remember the symmetry of the coefficients
$\binom{n}{r}=\binom{n}{n-r}$
Pascal's triangle is symmetric, so coefficients read the same forwards and backwards. This is a handy check that your numbers are right.
Check the number of terms
$(a+bx)^n \text{ has } n+1 \text{ terms}$
An expansion of power n always has one more term than the power. Counting the terms is a quick way to spot if you have dropped one.
Watch the powers add up to n
$\text{power of }a + \text{power of }bx = n$
In every single term the two powers must add to n. If they don't, an arithmetic slip has crept in.
Recall Pascal's triangle relation
$\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$
Each entry in Pascal's triangle is the sum of the two above it. This is where the coefficients come from without any factorials.
Note the end coefficients
$\binom{n}{0}=1,\qquad \binom{n}{n}=1$
The very first and very last coefficients are always 1. That gives you two terms of the answer for free.
Be careful with the signs
$(-1)^r>0 \text{ when } r \text{ is even, } <0 \text{ when } r \text{ is odd}$
When the bracket contains a minus sign the terms alternate in sign. Track this carefully so no negative is lost.
Recall the binomial theorem
$(a+b)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r$
This single formula generates every term of the expansion. Getting comfortable writing it down first stops us from missing terms later.
Recall the factorial form of the coefficient
$\binom{n}{r}=\frac{n!}{r!\,(n-r)!}$
The binomial coefficient just counts how many ways r objects can be chosen from n. On your calculator this is the nCr button.
Remember the symmetry of the coefficients
$\binom{n}{r}=\binom{n}{n-r}$
Pascal's triangle is symmetric, so coefficients read the same forwards and backwards. This is a handy check that your numbers are right.
Evaluate
$3^5=243$
Since 1+2=3, the answer is 3^5=243. Substituting a value into an expansion is a powerful trick.

Answer

243

Question 5

9 markschallenging

Find the coefficient of

x^{6}

in the expansion of

(1 + x)^{4}(1 - x)^{8}

Show worked solution

Worked solution

Plan the method
$\text{find terms whose powers multiply to give } x^{6}$
We do not need the full expansion of the product. We only need the pieces from each bracket whose x-powers add up to the power we want.
Expand (1 + x)^{4} as far as needed
$1 + 4x + 6x^{2} + 4x^{3} + x^{4}$
We expand the first bracket only up to the power we need. Any higher terms cannot contribute to our target term.
Expand (1 - x)^{8} as far as needed
$1 - 8x + 28x^{2} - 56x^{3} + 70x^{4} - 56x^{5} + 28x^{6}$
Same idea for the second bracket: stop once the powers get too big to help. This saves a lot of work.
Pick pairs of powers that add to 6
$x^{0}\cdot x^{6} ,\ x^{1}\cdot x^{5} ,\ x^{2}\cdot x^{4} ,\ x^{3}\cdot x^{3} ,\ x^{4}\cdot x^{2}$
To make x^{6} we multiply a term from the first bracket by a term from the second so the powers total 6. Every such pairing contributes.
Recall the binomial theorem
$(a+b)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r$
This single formula generates every term of the expansion. Getting comfortable writing it down first stops us from missing terms later.
Recall the factorial form of the coefficient
$\binom{n}{r}=\frac{n!}{r!\,(n-r)!}$
The binomial coefficient just counts how many ways r objects can be chosen from n. On your calculator this is the nCr button.
Remember the symmetry of the coefficients
$\binom{n}{r}=\binom{n}{n-r}$
Pascal's triangle is symmetric, so coefficients read the same forwards and backwards. This is a handy check that your numbers are right.
Check the number of terms
$(a+bx)^n \text{ has } n+1 \text{ terms}$
An expansion of power n always has one more term than the power. Counting the terms is a quick way to spot if you have dropped one.
Watch the powers add up to n
$\text{power of }a + \text{power of }bx = n$
In every single term the two powers must add to n. If they don't, an arithmetic slip has crept in.
Recall Pascal's triangle relation
$\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$
Each entry in Pascal's triangle is the sum of the two above it. This is where the coefficients come from without any factorials.
Note the end coefficients
$\binom{n}{0}=1,\qquad \binom{n}{n}=1$
The very first and very last coefficients are always 1. That gives you two terms of the answer for free.
Be careful with the signs
$(-1)^r>0 \text{ when } r \text{ is even, } <0 \text{ when } r \text{ is odd}$
When the bracket contains a minus sign the terms alternate in sign. Track this carefully so no negative is lost.
Recall the binomial theorem
$(a+b)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r$
This single formula generates every term of the expansion. Getting comfortable writing it down first stops us from missing terms later.
Recall the factorial form of the coefficient
$\binom{n}{r}=\frac{n!}{r!\,(n-r)!}$
The binomial coefficient just counts how many ways r objects can be chosen from n. On your calculator this is the nCr button.
Multiply the matching coefficients
$(1)(28) + (4)(-56) + (6)(70) + (4)(-56) + (1)(28) = 28$
We multiply the coefficients of each matching pair and add the products. The sum is the coefficient we are after.

Answer

28

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