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Proof by Induction
NOTE: Clicking an entry in the right column will take you directly to that question!
(BLUE = Paper 1, RED = Paper 2)
2021 Paper 1 Question 4(a)
(a) Prove using induction that \(2^{3n-1}\) is divisible by \(7\) for all \(n\in\mathbb{N}\).
The answer is already in the question!
When \(n=1\):
\begin{align}2^{3n-1}+3&=2^{3(1)-1}+3\\&=2^2+3\\&=7\end{align}
which is divisible by \(7\).
Therefore, let us assume that it is true for \(n=k\), i.e. that \(2^{3k-1}+3\) is divisible by \(7\).
Give that it is true for \(n=k\) , we wish to now prove that it is also true for \(n=k+1\), i.e. that \(2^{3(k+1)-1}+3\) is divisible by \(7\).
\begin{align}2^{3(k+1)-1}+3&=2^{3k+3-1}+3\\&=2^{3k+2}+3\\&=2^{3+(3k-1)}+3\\&=2^32^{3k-1}+3\\&=(8)2^{3k-1}+3\\&=(7+1)2^{3k-1}+3\\&=7^{3k-1}+(2^{3k-1}+3)\end{align}
As both terms are divisible by \(7\), \(2^{3(k+1)-1}+3\) is divisible by \(7\), as required.
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2020 Paper 1 Question 7(d)
(d) Prove using induction that, for all \(n\in\mathbb{N}\), the sum of the first \(n\) square numbers can be found using the formula:
\begin{align}1^2+2^2+3^2+4^2+…+n^2=\frac{n(n+1)(2n+1)}{6}\end{align}
The answer is already in the question!
When \(n=1\), the right hand side is
\begin{align}\frac{1(1+1)(2(1)+1)}{6}&=\frac{1(2)(3)}{6}\\&=1\end{align}
which is indeed equal to the left hand side of \(1^2\).
Therefore, let us assume that this relation is true for \(n=k\), i.e. that
\begin{align}1^2+2^2+3^2+4^2+…+k^2=\frac{k(k+1)(2k+1)}{6}\end{align}
Using this assumption, we now wish to prove that it is true for \(n=k+1\), i.e. that
\begin{align}1^2+2^2+3^2+4^2+…+k^2+(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}\end{align}
According to our assumption, we can rewrite the left hand side as
\begin{align}1^2+2^2+3^2+4^2+…+k^2+(k+1)^2&=\frac{k(k+1)(2k+1)}{6}+(k+1)^2\\&=\frac{k(k+1)(2k+1)+6(k+1)^2}{6}\\&=\frac{(k+1)[k(2k+1)+6(k+1)]}{6}\\&=\frac{(k+1)(2k^2+7k+6)}{6}\\&=\frac{(k+1)(k+2)(2k+3)}{6}\end{align}
which is indeed equal to the right hand side.
Therefore, the formula is indeed true for all \(n\in\mathbb{N}\).
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2019 Paper 1 Question 2(b)
(b) Prove, using induction, that \(3^n\geq 4n+1\), where \(n\geq2\) and \(n\in\mathbb{N}\).
The answer is already in the question!
The inequality is true for \(n=2\) as \(3^2\geq4(2)+1\).
Therefore let, us assume that it is true for \(n=k\), i.e. that
\begin{align}3^k\geq4k+1\end{align}
Using this assumption, we now wish to prove that it is also true for \(n=k+1\), i.e. that
\begin{align}3^{k+1}\geq4k+5\end{align}
First, we will multiply our assumption by \(3\):
\begin{align}3(3^k)\geq3(4k+1)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3^{k+1}\geq12k+3\end{align}
Since \(12k+3>4k+5\) for \(k\geq2\), it follows that
\begin{align}3^{k+1}\geq4k+5\end{align}
as required.
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2018 Paper 1 Question 4(a)
(a) Prove, using induction, that if \(n\) is a positive integer then
\((\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(n\theta)\), where \(i^2=-1\).
The answer is already in the question!
For \(n=1\), the formula states that
\begin{align}(\cos\theta+i\sin\theta)^1=\cos(1\theta)+i\sin(1\theta)\end{align}
\begin{align}\cos\theta+i\sin\theta=\cos\theta+i\sin\theta\end{align}
and therefore it is true for \(n=1\).
Hence, let us assume that it is true for \(n=k\), i.e. that
\begin{align}(\cos\theta+i\sin\theta)^k=\cos(k\theta)+i\sin(k\theta)\end{align}
Using this assumption, we now wish to prove that it is also true for \(n=k+1\), i.e. that
\begin{align}(\cos\theta+i\sin\theta)^{k+1}=\cos[(k+1)\theta]+i\sin[(k+1)\theta]\end{align}
Let us look at the left hand side:
\begin{align}(\cos\theta+i\sin\theta)^{k+1}&=(\cos\theta+i\sin\theta)^{k}(\cos\theta+i\sin\theta)^{1}\\&=[\cos(k\theta)+i\sin(k\theta)](\cos\theta+i\sin\theta)\theta]\\&=[\cos(k\theta)-\sin(k\theta)\sin\theta]+i[\cos(k\theta)\sin\theta+\cos\theta\sin(k\theta)]\\&=\cos[(k+1)\theta]+i\sin[(k+1)\theta]\end{align}
which is indeed the right hand side.
Therefore, the formula is true for all positive integers \(n\).
Our Video Walkthroughs are in the final stages of editing and will go live during Summer 2023.
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2016 Paper 1 Question 4(a)
(a) Prove by induction that \(8^n-1\) is divisible by \(7\) for all \(n\in\mathbb{N}\).
The answer is already in the question!
The statement is true for \(n=1\) as \(8^1-1=7\) is divisible by \(7\).
Therefore, let us assume that it is true for \(n=k\), i.e. that \(8^k-1\) is divisible by \(7\).
Using this assumption, we wish to now prove that \(8^{k+1}-1\) is also divisible by \(7\).
\begin{align}8^{k+1}-1&=8(8^k)-1\\&=(7+1)(8^k)-1\\&=7(8^k)+(8^k-1)\end{align}
As both terms are divisible by \(7\), \(8^{k+1}-1\) is divisible by \(7\), as required.
