Binomial expansion (positive integer n) – A-level Maths exam questions with mark scheme

1CoreCalculator4 marks

Find the first 4 terms, in ascending powers of $x$, of the binomial expansion of $$(2+3x)^{5},$$ giving each term in its simplest form. [4]

$32$B1
for the correct structure of at least one term, e.g. $\binom{5}{2}2^{3}(3x)^{2}$ – correct binomial coefficient with correct powers of 2 and $3x$M1
two of $240x,\ 720x^{2},\ 1080x^{3}$ correctA1
all of $32+240x+720x^{2}+1080x^{3}$A1
Answer: $32+240x+720x^{2}+1080x^{3}$
Examiner tip: Bracket the $(3x)$ – writing $3x^{2}$ instead of $(3x)^{2}=9x^{2}$ is the most common lost A mark.

2StandardCalculator4 marks

In the binomial expansion of $(1+kx)^{8}$, where $k$ is a positive constant, the coefficient of $x^{2}$ is $112$.

(a) Find the value of $k$. [2]

(b) Find the coefficient of $x^{3}$ in the expansion. [2]

(a) for $\binom82k^{2}=112$, i.e. $28k^{2}=112$M1
(a) $k=2$ onlyA1
(b) for $\binom83\times(\text{their }k)^{3}$M1
(b) $448$A1
Answer: (a) $k=2$ (b) $448$
Examiner tip: The coefficient is $28k^{2}$, not $28k$ – square the $k$ along with the $x$; and reject $k=-2$ because $k>0$.

3StandardCalculator7 marks

(a) Find the first 4 terms, in ascending powers of $x$, of the binomial expansion of $$\left(1-\frac{x}{2}\right)^{10},$$ giving each term in its simplest form. [4]

(b) Use your expansion, with a suitable value of $x$, to find an approximation for $0.995^{10}$. Give your answer to 5 decimal places. [3]

(a) $1-5x$B1
(a) for correct structure for the $x^{2}$ or $x^{3}$ term, e.g. $\dfrac{10\times9}{2!}\left(-\dfrac{x}{2}\right)^{2}$M1
(a) $+\tfrac{45}{4}x^{2}$ (allow $+11.25x^{2}$)A1
(a) $-15x^{3}$A1
(b) $x=0.01$ stated or usedB1
(b) for substituting their $x$ into their 4-term expansion (allow if $x=0.005$ used)M1
(b) $0.95111$A1
Answer: (a) $1-5x+\tfrac{45}{4}x^{2}-15x^{3}$ (b) $0.95111$
Examiner tip: Solve $1-\frac{x}{2}=0.995$ to get $x=0.01$; using $x=0.005$ is a common slip that loses the B1 and the final A1.

4HardCalculator4 marks

Find the term independent of $x$ in the binomial expansion of $$\left(2x+\frac{1}{x^{2}}\right)^{9}.$$ [4]

for a general term of the form $\binom9r(2x)^{9-r}\left(x^{-2}\right)^{r}$ (or equivalent)M1
for setting the power of $x$ to zero: $9-r-2r=0$M1
$r=3$ (or identifies the $\binom93$ term)A1
$5376$A1
Answer: $5376$
Examiner tip: Track powers of $x$ separately from the numbers: $(2x)^{6}$ contributes $2^{6}=64$, which students often forget.

5HardCalculator5 marks

(a) Find the first 3 terms, in ascending powers of $x$, of the binomial expansion of $(1-3x)^{6}$. [3]

(b) Hence find the coefficient of $x^{2}$ in the expansion of $$(2+x)(1-3x)^{6}.$$ [2]

(a) $1-18x$B1
(a) for $\binom62(-3x)^{2}$ or $\dfrac{6\times5}{2}(\pm3x)^{2}$M1
(a) $+135x^{2}$A1
(b) for $2\times135+1\times(-18)$ using their coefficients (two products)M1
(b) $252$A1
Answer: (a) $1-18x+135x^{2}$ (b) $252$
Examiner tip: In (b) there are two ways to make $x^{2}$: $2\times135x^{2}$ and $x\times(-18x)$ – don't forget the second.

6StretchCalculator7 marks

The binomial expansion of $(1+ax)^{n}$, where $n$ is a positive integer and $a$ is a constant, begins $$1+12x+63x^{2}+\dots$$

(a) Find the value of $n$ and the value of $a$. [5]

(b) Find the coefficient of $x^{3}$ in the expansion. [2]

(a) for $na=12$M1
(a) for $\dfrac{n(n-1)}{2}a^{2}=63$M1
(a) for a valid method to eliminate one variable, e.g. $\dfrac{n(n-1)}{2}\cdot\dfrac{144}{n^{2}}=63\Rightarrow 72(n-1)=63n$M1
(a) $n=8$A1
(a) $a=\tfrac32$A1
(b) for $\dfrac{n(n-1)(n-2)}{3!}a^{3}$ with their $n$ and $a$M1
(b) $189$A1
Answer: (a) $n=8,\ a=\tfrac32$ (b) $189$
Examiner tip: Square the $a$ in the $x^{2}$ coefficient; then substitute $a=\frac{12}{n}$ to get a linear equation in $n$.

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