CAIE FP1 2016 November — Question 4 6 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2016
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicBinomial Theorem (positive integer n)
TypeStandard binomial expansion
DifficultyChallenging +1.2 This is a two-part proof question requiring (1) algebraic manipulation of factorial expressions to verify Pascal's identity, and (2) a structured induction proof of the binomial theorem. While it requires formal proof technique and careful bookkeeping with the Pascal's identity result, both parts follow standard templates that Further Maths students practice extensively. The induction step is mechanical once Pascal's identity is established, making this moderately above average but not requiring novel insight.
Spec1.04a Binomial expansion: (a+b)^n for positive integer n4.01a Mathematical induction: construct proofs

4 Using factorials, show that \(\binom { n } { r - 1 } + \binom { n } { r } = \binom { n + 1 } { r }\). Hence prove by mathematical induction that $$( a + x ) ^ { n } = \binom { n } { 0 } a ^ { n } + \binom { n } { 1 } a ^ { n - 1 } x + \ldots + \binom { n } { r } a ^ { n - r } x ^ { r } + \ldots + \binom { n } { n } x ^ { n }$$ for every positive integer \(n\).

Question 4:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\binom{n}{r-1}+\binom{n}{r} = \frac{n!}{(r-1)!(n-r+1)!}+\frac{n!}{r!(n-r)!} = \frac{n!}{(r-1)!(n-r)!}\left(\frac{1}{n-r+1}+\frac{1}{r}\right)\)M1
\(= \frac{n!}{(r-1)!(n-r)!}\cdot\frac{r+n-r+1}{r(n-r+1)} = \frac{(n+1)!}{r!(n-r+1)!} = \binom{n+1}{r}\)A1 [2]
\((a+x)^1 = \binom{1}{0}a + \binom{1}{1}x = a+x \Rightarrow H_1\) is trueB1
Assume \(H_k\) is true: \((a+x)^k = \binom{k}{0}a^k+\binom{k}{1}a^{k-1}x+\ldots+\binom{k}{r}a^{k-r}x^r+\ldots+\binom{k}{k}x^k\)B1
Multiplying by \((a+x)\), the coefficient of \(a^{k-r+1}x^r\) is \(\binom{k}{r-1}+\binom{k}{r} = \binom{k+1}{r}\)M1
\(\Rightarrow H_{k+1}\) is true. Hence \(H_n\) is true for all positive integers.A1 [4]
## Question 4:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\binom{n}{r-1}+\binom{n}{r} = \frac{n!}{(r-1)!(n-r+1)!}+\frac{n!}{r!(n-r)!} = \frac{n!}{(r-1)!(n-r)!}\left(\frac{1}{n-r+1}+\frac{1}{r}\right)$ | M1 | |
| $= \frac{n!}{(r-1)!(n-r)!}\cdot\frac{r+n-r+1}{r(n-r+1)} = \frac{(n+1)!}{r!(n-r+1)!} = \binom{n+1}{r}$ | A1 | [2] |
| $(a+x)^1 = \binom{1}{0}a + \binom{1}{1}x = a+x \Rightarrow H_1$ is true | B1 | |
| Assume $H_k$ is true: $(a+x)^k = \binom{k}{0}a^k+\binom{k}{1}a^{k-1}x+\ldots+\binom{k}{r}a^{k-r}x^r+\ldots+\binom{k}{k}x^k$ | B1 | |
| Multiplying by $(a+x)$, the coefficient of $a^{k-r+1}x^r$ is $\binom{k}{r-1}+\binom{k}{r} = \binom{k+1}{r}$ | M1 | |
| $\Rightarrow H_{k+1}$ is true. Hence $H_n$ is true for all positive integers. | A1 | [4] |
4 Using factorials, show that $\binom { n } { r - 1 } + \binom { n } { r } = \binom { n + 1 } { r }$.

Hence prove by mathematical induction that

$$( a + x ) ^ { n } = \binom { n } { 0 } a ^ { n } + \binom { n } { 1 } a ^ { n - 1 } x + \ldots + \binom { n } { r } a ^ { n - r } x ^ { r } + \ldots + \binom { n } { n } x ^ { n }$$

for every positive integer $n$.

\hfill \mbox{\textit{CAIE FP1 2016 Q4 [6]}}