Question 2 - A-Level Maths

CAIE Further Paper 1 2023 November — Question 2 6 marks

Exam Board	CAIE
Module	Further Paper 1 (Further Paper 1)
Year	2023
Session	November
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Proof by induction
Type	Prove summation with exponentials
Difficulty	Standard +0.8 This is a non-trivial induction proof requiring careful algebraic manipulation of the inductive step. The formula involves exponentials and a squared denominator, making the algebra more complex than standard summation proofs. However, it follows a standard induction structure without requiring novel insight, placing it moderately above average difficulty for Further Maths students.
Spec	4.01a Mathematical induction: construct proofs

2 Prove by mathematical induction that, for all positive integers $n$, $$1 + 2 x + 3 x ^ { 2 } + \ldots + n x ^ { n - 1 } = \frac { 1 - ( n + 1 ) x ^ { n } + n x ^ { n + 1 } } { ( 1 - x ) ^ { 2 } }$$

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Question 2:

Answer	Marks	Guidance
Answer	Marks	Guidance
$1 = \frac{1-2x+x^2}{(1-x)^2} = \frac{(1-x)^2}{(1-x)^2}$ so $H_1$ is true	B1	Checks base case
Assume that $\sum_{r=1}^{k} rx^{r-1} = \frac{1-(k+1)x^k+kx^{k+1}}{(1-x)^2}$	B1	States inductive hypothesis
$\sum_{r=1}^{k+1} rx^{r-1} = \frac{1-(k+1)x^k+kx^{k+1}}{(1-x)^2} + (k+1)x^k$	M1	Considers sum to $k+1$
$\frac{1-(k+1)x^k+kx^{k+1}+(k+1)x^k(1-2x+x^2)}{(1-x)^2}$	M1	Puts over a common denominator
$\frac{1+kx^{k+1}+(k+1)x^k(-2x+x^2)}{(1-x)^2} = \frac{1-(k+2)x^{k+1}+(k+1)x^{k+2}}{(1-x)^2}$	A1
So $H_{k+1}$ is true. By induction, $H_n$ is true for all positive integers $n$	A1	States conclusion
Total: 6

## Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 = \frac{1-2x+x^2}{(1-x)^2} = \frac{(1-x)^2}{(1-x)^2}$ so $H_1$ is true | B1 | Checks base case |
| Assume that $\sum_{r=1}^{k} rx^{r-1} = \frac{1-(k+1)x^k+kx^{k+1}}{(1-x)^2}$ | B1 | States inductive hypothesis |
| $\sum_{r=1}^{k+1} rx^{r-1} = \frac{1-(k+1)x^k+kx^{k+1}}{(1-x)^2} + (k+1)x^k$ | M1 | Considers sum to $k+1$ |
| $\frac{1-(k+1)x^k+kx^{k+1}+(k+1)x^k(1-2x+x^2)}{(1-x)^2}$ | M1 | Puts over a common denominator |
| $\frac{1+kx^{k+1}+(k+1)x^k(-2x+x^2)}{(1-x)^2} = \frac{1-(k+2)x^{k+1}+(k+1)x^{k+2}}{(1-x)^2}$ | A1 | |
| So $H_{k+1}$ is true. By induction, $H_n$ is true for all positive integers $n$ | A1 | States conclusion |
| **Total: 6** | | |

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Show LaTeX source

2 Prove by mathematical induction that, for all positive integers $n$,

$$1 + 2 x + 3 x ^ { 2 } + \ldots + n x ^ { n - 1 } = \frac { 1 - ( n + 1 ) x ^ { n } + n x ^ { n + 1 } } { ( 1 - x ) ^ { 2 } }$$

\hfill \mbox{\textit{CAIE Further Paper 1 2023 Q2 [6]}}

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Q1 7 Q2 6 Q3 8 Q4 9 Q5 15 Q6 15 Q7 15