First, find the first derivative:

\(\frac{d}{dx}(x \ln x) = 1 + \ln x\).

Check the base case for \(n = 2\):

\(\frac{d^2}{dx^2}(x \ln x) = x^{-1} = (-1)^2 (2-2)! x^{1-2}\).

Assume that \(\frac{d^k}{dx^k}(x \ln x) = (-1)^k (k-2)! x^{1-k}\) holds for some integer \(k \geq 2\).

Differentiate the \(k\)-th derivative:

\(\frac{d^{k+1}}{dx^{k+1}}(x \ln x) = (-1)^k (k-2)!(1-k)x^{1-k-1}\).

Simplify:

\(= (-1)^{k+1} ((k+1)-2)! x^{1-(k+1)}\).

Thus, \(\frac{d^n}{dx^n}(x \ln x) = (-1)^n (n-2)! x^{1-n}\) is also true for \(n = k+1\).

By induction, the statement is true for every integer \(n \geq 2\).

Base case: For \(n = 1\),

\(\frac{d}{dx}(\arctan x) = \frac{1}{1+x^2},\)

which is true.

Inductive step: Assume

\(\frac{d^k}{dx^k}(\arctan x) = P_k(x)(1+x^2)^{-k},\)

where \(\deg P_k(x) = k-1\).

Then,

\(\frac{d^{k+1}}{dx^{k+1}}(\arctan x) = P_k'(x)(1+x^2)^{-k} - 2kxP_k(x)(1+x^2)^{-k-1}.\)

This can be written as

\(\left(P_k'(x)(1+x^2) - 2kxP_k(x)\right)(1+x^2)^{-k-1},\)

which implies \(\deg P_{k+1}(x) = k\).

Thus, the statement is true for \(n = k+1\). By induction, it is true for all positive integers \(n\).

Base case: For \(n = 1\), \(u_1 = 4 = 3^1 + 1\). This holds true.

Inductive step: Assume it is true for \(n = k\), so \(u_k = 3^k + 1\).

Then \(u_{k+1} = 3u_k - 2 = 3(3^k + 1) - 2 = 3^{k+1} + 3 - 2 = 3^{k+1} + 1\).

Thus, it is true for \(n = k+1\). Hence, by induction, \(u_n = 3^n + 1\) for all positive integers \(n\).

Base case: For \(n = 1\),

\(\frac{d}{dx} \left( \arctan x \right) = \frac{1}{1 + x^2},\)

which is true.

Inductive step: Assume

\(\frac{d^k}{dx^k} \left( \arctan x \right) = P_k(x) (1 + x^2)^{-k},\)

where \(\deg P_k(x) = k - 1\).

Then,

\(\frac{d^{k+1}}{dx^{k+1}} \left( \arctan x \right) = P_k'(x) (1 + x^2)^{-k} - 2kx P_k(x) (1 + x^2)^{-k-1}.\)

This can be written as

\(\left( P_k'(x) (1 + x^2) - 2kx P_k(x) \right) (1 + x^2)^{-k-1},\)

so \(\deg P_{k+1}(x) = k\).

Thus, true for \(n = k + 1\). By induction, true for all positive integers \(n\).

(a) Base case: For \(n = 1\), \(\sum_{r=1}^{1} r^2 = 1^2 = 1\) and \(\frac{1}{6} \times 1 \times 2 \times 3 = 1\). So \(H_1\) is true.

Inductive step: Assume \(\sum_{r=1}^{k} r^2 = \frac{1}{6}k(k+1)(2k+1)\) is true for \(n = k\).

Consider \(\sum_{r=1}^{k+1} r^2 = \sum_{r=1}^{k} r^2 + (k+1)^2\).

\(= \frac{1}{6}k(k+1)(2k+1) + (k+1)^2\)

\(= \frac{1}{6}(k+1)(2k^2 + k + 6k + 6)\)

\(= \frac{1}{6}(k+1)(2k^2 + 7k + 6)\)

\(= \frac{1}{6}(k+1)(k+2)(2k+3)\)

So \(H_{k+1}\) is true. By induction, \(H_n\) is true for all positive integers \(n\).

(b) Using the identity, expand and simplify:

\((2n+1)^5 - 1\)

\(= 160S_n + \frac{40}{3}n(n+1)(2n+1) + 2n\)

\(160S_n = (2n+1)^5 - \frac{40}{3}n(n+1)(2n+1) - (2n+1)\)

\(160S_n = (2n+1)((2n+1)^4 - \frac{40}{3}n(n+1) - 1)\)

\(\Rightarrow S_n = \frac{1}{30}n(n+1)(2n+1)(3n^2 + 3n - 1)\)

(c) \(n^{-5}S_n = \frac{1}{30}(1 + \frac{1}{n})(2 + \frac{1}{n})(3 + \frac{3}{n} - \frac{1}{n^2})\)

As \(n \to \infty\), \(n^{-5}S_n \to \frac{1}{5}\).

Base case: For \(n = 1\),

\(1 = \frac{1 - 2x + x^2}{(1-x)^2} = \frac{(1-x)^2}{(1-x)^2}.\)

So \(H_1\) is true.

Inductive step: Assume \(\sum_{r=1}^{k} rx^{r-1} = \frac{1 - (k+1)x^k + kx^{k+1}}{(1-x)^2}\).

Consider \(\sum_{r=1}^{k+1} rx^{r-1} = \frac{1 - (k+1)x^k + kx^{k+1}}{(1-x)^2} + (k+1)x^k\).

\(= \frac{1 - (k+1)x^k + kx^{k+1} + (k+1)x^k(1-2x+x^2)}{(1-x)^2}\)

\(= \frac{1 + kx^{k+1} + (k+1)x^k(-2x+x^2)}{(1-x)^2}\)

\(= \frac{1 - (k+2)x^{k+1} + (k+1)x^{k+2}}{(1-x)^2}.\)

So \(H_{k+1}\) is true. By induction, \(H_n\) is true for all positive integers \(n\).

Base case: For \(n = 1\),

\(\frac{d}{dx} \left( x^2 e^x \right) = x^2 e^x + 2x e^x = (x^2 + 2x) e^x.\)

So true when \(n = 1\).

Inductive step: Assume that

\(\frac{d^k}{dx^k} \left( x^2 e^x \right) = \left( x^2 + 2kx + k(k-1) \right) e^x\)

for some value of \(k\).

Then,

\(\frac{d^{k+1}}{dx^{k+1}} \left( x^2 e^x \right) = \left( x^2 + 2kx + k(k-1) \right) e^x + e^x (2x + 2k).\)

Simplifying gives:

\(\left( x^2 + 2(k+1)x + k(k+1) \right) e^x.\)

So true when \(n = k+1\). By induction, true for all positive integers \(n\).

Base case: For \(n = 1\),

\(1 = \frac{1 - 2x + x^2}{(1-x)^2} = \frac{(1-x)^2}{(1-x)^2},\)

so \(H_1\) is true.

Inductive step: Assume that

\(\sum_{r=1}^{k} r x^{r-1} = \frac{1 - (k+1)x^k + kx^{k+1}}{(1-x)^2}.\)

Consider \(\sum_{r=1}^{k+1} r x^{r-1}\):

\(\sum_{r=1}^{k+1} r x^{r-1} = \frac{1 - (k+1)x^k + kx^{k+1}}{(1-x)^2} + (k+1)x^k.\)

Combine over a common denominator:

\(\frac{1 - (k+1)x^k + kx^{k+1} + (k+1)x^k(1-2x+x^2)}{(1-x)^2}.\)

Simplify:

\(\frac{1 + kx^{k+1} + (k+1)x^k(-2x + x^2)}{(1-x)^2} = \frac{1 - (k+2)x^{k+1} + (k+1)x^{k+2}}{(1-x)^2}.\)

Thus, \(H_{k+1}\) is true. By induction, \(H_n\) is true for all positive integers \(n\).

(a) Base case: For \(n = 1\),

\(2\mathbf{A} = \begin{pmatrix} 6 & 0 \\ 2 & 2 \end{pmatrix} = \begin{pmatrix} 2 \times 3^1 & 0 \\ 3^1 - 1 & 2 \end{pmatrix}.\)

Assume true for \(n = k\), so

\(2\mathbf{A}^k = \begin{pmatrix} 2 \times 3^k & 0 \\ 3^k - 1 & 2 \end{pmatrix}.\)

Then for \(n = k + 1\),

\(2\mathbf{A}^{k+1} = \begin{pmatrix} 2 \times 3^{k+1} & 0 \\ 3^{k+1} - 3 + 2 & 2 \end{pmatrix}.\)

Thus, true for \(n = k + 1\). By induction, true for all positive integers.

(b) \(\det \mathbf{A}^n = \det \begin{pmatrix} 3^n & 0 \\ \frac{1}{2}(3^n - 1) & 1 \end{pmatrix} = 3^n\).

\(\mathbf{A}^{-n} = 3^{-n} \begin{pmatrix} 1 & 0 \\ \frac{1}{2}(1 - 3^n) & 3^n \end{pmatrix}\).

(a) Base case: For \(n = 1\),

\(2A = \begin{pmatrix} 6 & 0 \\ 2 & 2 \end{pmatrix} = \begin{pmatrix} 2 \times 3^1 & 0 \\ 3^1 - 1 & 2 \end{pmatrix}.\)

Assume true for \(n = k\), so \(2A^k = \begin{pmatrix} 2 \times 3^k & 0 \\ 3^k - 1 & 2 \end{pmatrix}\).

Then for \(n = k+1\),

\(2A^{k+1} = \begin{pmatrix} 2 \times 3^k & 0 \\ 3^k - 1 & 2 \end{pmatrix} \begin{pmatrix} 3 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 2 \times 3^{k+1} & 0 \\ 3^{k+1} - 3 + 2 & 2 \end{pmatrix}.\)

Thus, true for \(n = k+1\). By induction, true for all positive integers.

(b) \(\det A^n = \det \begin{pmatrix} 3^n & 0 \\ \frac{1}{2}(3^n - 1) & 1 \end{pmatrix} = 3^n\).

\(A^{-n} = 3^{-n} \begin{pmatrix} 1 & 0 \\ \frac{1}{2}(1 - 3^n) & 3^n \end{pmatrix}\).

(a) Base case: \(u_1 > 4\) (given).

Inductive step: Assume \(u_k > 4\) for some positive integer \(k\).

Then \(u_{k+1} - 4 = \frac{u_k^2 + u_k + 12}{2u_k} - 4 = \frac{u_k^2 - 7u_k + 12}{2u_k}\).

\(u_{k+1} - 4 = \frac{(u_k - 3)(u_k - 4)}{2u_k} > 0\).

Hence, by induction, \(u_n > 4\) for all positive integers \(n\).

(b) \(u_{n+1} - u_n = \frac{-u_n^2 + u_n + 12}{2u_n} = \frac{-(u_n - 4)(u_n + 3)}{2u_n}\).

So \(u_n > 4 \Rightarrow u_{n+1} - u_n < 0\).

(a) The matrix \(\mathbf{A} = \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}\) represents a shear in the \(x\)-direction.

(b) Base case: For \(n = 1\), \(\mathbf{A}^1 = \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}\), which is true.

Inductive step: Assume true for \(n = k\), so \(\mathbf{A}^k = \begin{pmatrix} 1 & ka \\ 0 & 1 \end{pmatrix}\).

Then \(\mathbf{A}^{k+1} = \mathbf{A}^k \mathbf{A} = \begin{pmatrix} 1 & ka \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & a + ak \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & (k+1)a \\ 0 & 1 \end{pmatrix}\).

Thus, by induction, \(\mathbf{A}^n = \begin{pmatrix} 1 & na \\ 0 & 1 \end{pmatrix}\) for all positive integers \(n\).

(c) \(\mathbf{A}^n \mathbf{B} = \begin{pmatrix} 1 & na \\ 0 & 1 \end{pmatrix} \begin{pmatrix} b & b \\ a^{-1} & a^{-1} \end{pmatrix} = \begin{pmatrix} b+n & b+n \\ \frac{1}{a} & \frac{1}{a} \end{pmatrix}\).

Transform \(\begin{pmatrix} x \\ y \end{pmatrix}\) by multiplying matrices to find \(\begin{pmatrix} X \\ Y \end{pmatrix}\).

\(\begin{pmatrix} b+n & b+n \\ \frac{1}{a} & \frac{1}{a} \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} (b+n)(x+y) \\ \frac{1}{a}(x+y) \end{pmatrix}\).

\(\frac{1}{a}(1+m) = m(b+n)(1+m)\)

\(y = -x\)

\(y = \frac{1}{a(b+n)}x\)

Base case: For \(n = 1\),

\(\frac{d}{dx}(xf(x)) = xf'(x) + f(x) = xf'(x) + (2(1)-1)f(x).\)

Assume true for \(n = k\), so

\(\frac{d^{2k-1}}{dx^{2k-1}}(xf(x)) = xf'(x) + (2k-1)f(x).\)

Then,

\(\frac{d^{2k}}{dx^{2k}}(xf(x)) = xf(x) + 2kf'(x).\)

\(\frac{d^{2k+1}}{dx^{2k+1}}(xf(x)) = xf'(x) + f(x) + 2kf(x)\)

\(= xf'(x) + (2k+1)f(x).\)

So, it is also true for \(n = k+1\). Hence, by induction, true for all positive integers.

(a) Base case: For \(n = 1\),

\(5 \times 1^4 + 1^2 = \frac{1}{2} (2)^2 (2+1) = 6\), so \(H_1\) is true.

Inductive step: Assume \(\sum_{r=1}^{k} (5r^4 + r^2) = \frac{1}{2} k^2 (k+1)^2 (2k+1)\).

Consider \(\sum_{r=1}^{k+1} (5r^4 + r^2) = \frac{1}{2} k^2 (k+1)^2 (2k+1) + 5(k+1)^4 + (k+1)^2\).

\(= \frac{1}{2} (k+1)^2 (2k^3 + k^2 + 10(k+1)^2 + 2)\)

\(= \frac{1}{2} (k+1)^2 (k^2 + 2k + 1)(2k+3)\)

So \(H_{k+1}\) is true. By induction, \(H_n\) is true for all positive integers \(n\).

(b) Using the result from part (a):

\(5 \sum_{r=1}^{n} r^4 + \frac{1}{6} n(n+1)(2n+1) = \frac{1}{2} n^2 (n+1)^2 (2n+1)\)

\(5 \sum_{r=1}^{n} r^4 = \frac{1}{6} n(n+1)(2n+1)(3n(n+1)-1)\)

\(\sum_{r=1}^{n} r^4 = \frac{1}{30} n(n+1)(2n+1)(3n+3n-1)\)

(a) Base case: For \(n = 2\), \(a_2 = 2^3\) leading to \(\ln a_2 = 3 \ln 2\), so true when \(n = 2\).

Inductive step: Assume \(\ln a_k \geq 3^{k-1} \ln 2\).

Then \(\ln a_{k+1} = 3 \ln \left( a_k + \frac{1}{a_k} \right) > 3 \ln a_k \geq 3^k \ln 2\).

Thus, true for \(n = k+1\). By induction, true for all integers \(n \geq 2\).

(b) \(3 \ln \left( a_n + \frac{1}{a_n} \right) - \ln a_n > 2 \ln a_n\)

\(> 2 \times 3^{n-1} \ln 2 = 3^{n-1} \ln 4\)

First, differentiate \(y = xe^{ax}\) once using the product rule:

\(\frac{dy}{dx} = axe^{ax} + e^{ax} = (ax + 1)e^{ax}\), so true when \(n = 1\).

Assume that \(\frac{d^k y}{dx^k} = \left( a^k x + ka^{k-1} \right) e^{ax}\).

Differentiate the \(k\)-th derivative:

\(\frac{d^{k+1} y}{dx^{k+1}} = a \left( a^k x + ka^{k-1} \right) e^{ax} + e^{ax} \left( a^k \right) = \left( a^{k+1} + (k+1)a^k \right) e^{ax}\).

So true when \(n = k + 1\). By induction, true for all positive integers \(n\).

Answer: For every positive integer \(n\), \(\dfrac{\mathrm d^{2n-1}}{\mathrm d x^{2n-1}}(x\sin x)=(-1)^{n-1}\bigl(x\cos x+(2n-1)\sin x\bigr)\).

We prove the result by mathematical induction.

Base case: Let \(n=1\). Then \(2n-1=1\), and, using the product rule,

\(\dfrac{\mathrm d}{\mathrm dx}(x\sin x)=x\cos x+\sin x\).

The right hand side of the proposed formula is

\((-1)^0\bigl(x\cos x+(2(1)-1)\sin x\bigr)=x\cos x+\sin x\).

So the formula is true for \(n=1\).

Inductive step: Assume that the formula is true for some positive integer \(k\). That is, assume

\(\dfrac{\mathrm d^{2k-1}}{\mathrm d x^{2k-1}}(x\sin x)=(-1)^{k-1}\bigl(x\cos x+(2k-1)\sin x\bigr)\).

Differentiate both sides once:

\(\dfrac{\mathrm d^{2k}}{\mathrm d x^{2k}}(x\sin x)=(-1)^{k-1}\dfrac{\mathrm d}{\mathrm dx}\bigl(x\cos x+(2k-1)\sin x\bigr)\).

Now \(\dfrac{\mathrm d}{\mathrm dx}(x\cos x)=\cos x-x\sin x\), and \(\dfrac{\mathrm d}{\mathrm dx}((2k-1)\sin x)=(2k-1)\cos x\). Hence

\(\dfrac{\mathrm d^{2k}}{\mathrm d x^{2k}}(x\sin x)=(-1)^{k-1}\bigl(-x\sin x+2k\cos x\bigr)\).

Differentiate once more:

\(\dfrac{\mathrm d^{2k+1}}{\mathrm d x^{2k+1}}(x\sin x)=(-1)^{k-1}\dfrac{\mathrm d}{\mathrm dx}\bigl(-x\sin x+2k\cos x\bigr)\).

Since \(\dfrac{\mathrm d}{\mathrm dx}(-x\sin x)=-\sin x-x\cos x\), and \(\dfrac{\mathrm d}{\mathrm dx}(2k\cos x)=-2k\sin x\), we get

\(\dfrac{\mathrm d^{2k+1}}{\mathrm d x^{2k+1}}(x\sin x)=(-1)^{k-1}\bigl(-x\cos x-(2k+1)\sin x\bigr)\).

Therefore

\(\dfrac{\mathrm d^{2k+1}}{\mathrm d x^{2k+1}}(x\sin x)=(-1)^k\bigl(x\cos x+(2k+1)\sin x\bigr)\).

But this is exactly the required formula with \(n=k+1\), since \(2(k+1)-1=2k+1\).

Thus, if the formula is true for \(n=k\), it is true for \(n=k+1\). Since it is true for \(n=1\), it follows by induction that it is true for every positive integer \(n\).

Answer:

1. (a) \(d=15\).
2. (b) \(\displaystyle \mathbf A^n=\begin{pmatrix}2^n&0\\2^n-1&1\end{pmatrix}\) for all positive integers \(n\).
3. (c) \(n=5\).

(a) The area scale factor of a matrix transformation is the absolute value of its determinant. For \(\mathbf A=\begin{pmatrix}2&0\\1&1\end{pmatrix}\),

\(\det \mathbf A=2\), so \(\det(\mathbf A^{-1})=(\det \mathbf A)^{-1}=\frac{1}{2}\).

Therefore the image area is \(30\times \frac{1}{2}=15\), so \(d=15\).

(b) We prove by induction that

\(\displaystyle \mathbf A^n=\begin{pmatrix}2^n&0\\2^n-1&1\end{pmatrix}\).

For \(n=1\),

\(\displaystyle \mathbf A=\begin{pmatrix}2&0\\1&1\end{pmatrix}=\begin{pmatrix}2^1&0\\2^1-1&1\end{pmatrix}\),

so the result is true when \(n=1\).

Assume it is true for \(n=k\), where \(k\) is a positive integer. Then

\(\displaystyle \mathbf A^k=\begin{pmatrix}2^k&0\\2^k-1&1\end{pmatrix}\).

Multiplying by \(\mathbf A\),

\(\displaystyle \mathbf A^{k+1}=\mathbf A^k\mathbf A=\begin{pmatrix}2^k&0\\2^k-1&1\end{pmatrix}\begin{pmatrix}2&0\\1&1\end{pmatrix}=\begin{pmatrix}2^{k+1}&0\\2(2^k-1)+1&1\end{pmatrix}\).

Since \(2(2^k-1)+1=2^{k+1}-1\),

\(\displaystyle \mathbf A^{k+1}=\begin{pmatrix}2^{k+1}&0\\2^{k+1}-1&1\end{pmatrix}\).

So, by induction, the formula is true for all positive integers \(n\).

(c) Using part (b),

\(\displaystyle \mathbf A^n\mathbf B=\begin{pmatrix}2^n&0\\2^n-1&1\end{pmatrix}\begin{pmatrix}1&0\\33&0\end{pmatrix}=\begin{pmatrix}2^n&0\\2^n+32&0\end{pmatrix}\).

A point \((x,y)\) is transformed to

\(\displaystyle \begin{pmatrix}2^n&0\\2^n+32&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}2^n x\\(2^n+32)x\end{pmatrix}\).

For the line \(y=2x\) to be invariant, the image must also satisfy \(y=2x\). Hence

\((2^n+32)x=2(2^n x)\).

For non-zero points on the line, this gives \(2^n+32=2^{n+1}\), so \(2^n=32\). Therefore \(n=5\).

Answer: (i) For every positive integer \(n\),

\(1+z+z^2+\cdots+z^{n-1}=\dfrac{z^n-1}{z-1}\) for \(z eq 1\).

(ii) Hence

\(\displaystyle \sum_{m=1}^{\infty}\left(\frac12\right)^m\sin m\theta=\frac{2\sin\theta}{5-4\cos\theta}.\)

(i) We prove the result by induction on \(n\).

Let \(P(n)\) be the statement

\(1+z+z^2+\cdots+z^{n-1}=\dfrac{z^n-1}{z-1}\), where \(z eq 1\).

Base case: When \(n=1\), the left-hand side is just \(1\). The right-hand side is

\(\dfrac{z^1-1}{z-1}=\dfrac{z-1}{z-1}=1\).

So \(P(1)\) is true.

Inductive step: Assume \(P(k)\) is true for some positive integer \(k\), so

\(1+z+z^2+\cdots+z^{k-1}=\dfrac{z^k-1}{z-1}.\)

Then for \(k+1\),

\(1+z+z^2+\cdots+z^{k-1}+z^k = \dfrac{z^k-1}{z-1}+z^k.\)

Putting this over a common denominator gives

\(\dfrac{z^k-1}{z-1}+z^k = \dfrac{z^k-1+z^k(z-1)}{z-1} = \dfrac{z^{k+1}-1}{z-1}.\)

Thus \(P(k+1)\) is true whenever \(P(k)\) is true.

Therefore, by mathematical induction,

\(1+z+z^2+\cdots+z^{n-1}=\dfrac{z^n-1}{z-1}\)

for all positive integers \(n\), provided \(z eq 1\).

(ii) Let

\(z=\dfrac12(\cos\theta+i\sin\theta)=\dfrac12 e^{i\theta}.\)

Since \(|z|=\frac12<1\), we can use the geometric-series result from part (i) in the limit as \(n\to\infty\):

\(\sum_{m=0}^{\infty} z^m = \dfrac{1}{1-z}.\)

Therefore

\(\sum_{m=1}^{\infty} z^m = \dfrac{z}{1-z}.\)

Now take imaginary parts. Since \(z^m=\left(\frac12\right)^m(\cos m\theta+i\sin m\theta)\), the imaginary part of \(\sum_{m=1}^{\infty} z^m\) is exactly \(\sum_{m=1}^{\infty}\left(\frac12\right)^m\sin m\theta\).

So we compute

\(\sum_{m=1}^{\infty} z^m = \frac{z}{1-z} = \frac{\frac12(\cos\theta+i\sin\theta)}{1-\frac12(\cos\theta+i\sin\theta)}.\)

Multiply top and bottom by \(2\):

\(\frac{\cos\theta+i\sin\theta}{2-\cos\theta-i\sin\theta}.\)

Now multiply numerator and denominator by the conjugate \(2-\cos\theta+i\sin\theta\):

\(\frac{(\cos\theta+i\sin\theta)(2-\cos\theta+i\sin\theta)}{(2-\cos\theta)^2+\sin^2\theta}.\)

The denominator simplifies to

\((2-\cos\theta)^2+\sin^2\theta = 4-4\cos\theta+\cos^2\theta+\sin^2\theta = 5-4\cos\theta.\)

The numerator expands to

\(\cos\theta(2-\cos\theta)-\sin^2\theta + i\big(\sin\theta(2-\cos\theta)+\cos\theta\sin\theta\big).\)

The imaginary part is

\(2\sin\theta.\)

Hence

\(\sum_{m=1}^{\infty}\left(\frac12\right)^m\sin m\theta = \frac{2\sin\theta}{5-4\cos\theta}.\)

Answer: For every positive integer \(n\),

\(\displaystyle \frac{\mathrm d^n y}{\mathrm d x^n}=(-1)^{n-1}\frac{(n-1)!a^n}{(ax+1)^n}\).

We prove the formula by induction on \(n\).

Let \(P(n)\) be the statement

\(\displaystyle \frac{\mathrm d^n y}{\mathrm d x^n}=(-1)^{n-1}\frac{(n-1)!a^n}{(ax+1)^n}\), where \(y=\ln(ax+1)\).

Base case: \(n=1\).

Differentiate \(y=\ln(ax+1)\):

\(\displaystyle \frac{\mathrm dy}{\mathrm dx}=\frac{a}{ax+1}.\)

Now the formula for \(n=1\) gives

\(\displaystyle (-1)^{0}\frac{0!\,a^1}{(ax+1)^1}=\frac{a}{ax+1},\)

which matches. So \(P(1)\) is true.

Inductive step.

Assume \(P(k)\) is true for some positive integer \(k\), so

\(\displaystyle \frac{\mathrm d^k y}{\mathrm d x^k}=(-1)^{k-1}\frac{(k-1)!a^k}{(ax+1)^k}.\)

Differentiate both sides with respect to \(x\):

\(\displaystyle \frac{\mathrm d^{k+1}y}{\mathrm d x^{k+1}} = (-1)^{k-1}(k-1)!a^k \frac{\mathrm d}{\mathrm dx}\big((ax+1)^{-k}\big).\)

Using the chain rule,

\(\displaystyle \frac{\mathrm d}{\mathrm dx}\big((ax+1)^{-k}\big) = -k(ax+1)^{-k-1}\cdot a.\)

Hence

\(\displaystyle \frac{\mathrm d^{k+1}y}{\mathrm d x^{k+1}} = (-1)^{k-1}(k-1)!a^k\big(-ka(ax+1)^{-k-1}\big).\)

Simplifying,

\(\displaystyle \frac{\mathrm d^{k+1}y}{\mathrm d x^{k+1}} = (-1)^k k! a^{k+1}(ax+1)^{-(k+1)}.\)

So

\(\displaystyle \frac{\mathrm d^{k+1}y}{\mathrm d x^{k+1}} = (-1)^k\frac{k!a^{k+1}}{(ax+1)^{k+1}},\)

which is exactly the required formula with \(n=k+1\).

Therefore, by mathematical induction, for every positive integer \(n\),

\(\displaystyle \frac{\mathrm d^{n} y}{\mathrm d x^{n}}=(-1)^{n-1} \frac{(n-1)!a^{n}}{(ax+1)^{n}}.\)

Answer: (i) \(u_n=4\left(\dfrac54\right)^n+3\).

(ii) The series is convergent for \(-\dfrac45\lt x\lt\dfrac45\).

(iii) \(a=\dfrac{\sqrt5}{2}\) and \(b=2\sqrt5\).

First prove the formula by induction.

For \(n=1\),

\(4\left(\dfrac54\right)^1+3=5+3=8=u_1,\)

so the result is true for \(n=1\).

Assume that for some positive integer \(k\),

\(u_k=4\left(\dfrac54\right)^k+3.\)

Then

\(u_{k+1}=\dfrac{5u_k-3}{4}=\dfrac{5\left(4\left(\dfrac54\right)^k+3\right)-3}{4}.\)

Simplifying,

\(u_{k+1}=\dfrac{20\left(\dfrac54\right)^k+12}{4}=5\left(\dfrac54\right)^k+3=4\left(\dfrac54\right)^{k+1}+3.\)

Therefore the formula holds for all positive integers \(n\).

Now

\(u_r-3=4\left(\dfrac54\right)^r.\)

The infinite series is

\(\sum_{r=1}^{\infty}(u_r-3)x^r=\sum_{r=1}^{\infty}4\left(\dfrac54\right)^r x^r=\sum_{r=1}^{\infty}4\left(\dfrac{5x}{4}\right)^r.\)

This is geometric and converges when

\(\left|\dfrac{5x}{4}\right|\lt 1,\)

so

\(-\dfrac45\lt x\lt\dfrac45.\)

For the logarithmic sum,

\(\ln(u_n-3)=\ln\left(4\left(\dfrac54\right)^n\right)=\ln4+n\ln\dfrac54.\)

Hence

\(\sum_{n=1}^{N}\ln(u_n-3)=N\ln4+\left(\sum_{n=1}^{N}n\right)\ln\dfrac54.\)

Since \(\sum_{n=1}^{N}n=\dfrac{N(N+1)}{2}\),

\(\sum_{n=1}^{N}\ln(u_n-3)=\dfrac{N^2}{2}\ln\dfrac54+N\left(\ln4+\dfrac12\ln\dfrac54\right).\)

So

\(N^2\ln a=\dfrac{N^2}{2}\ln\dfrac54\), giving \(a=\sqrt{\dfrac54}=\dfrac{\sqrt5}{2}\),

and

\(N\ln b=N\ln\left(4\sqrt{\dfrac54}\right)\), giving \(b=4\cdot\dfrac{\sqrt5}{2}=2\sqrt5\).

Answer: The identity is true for all positive integers \(n\).

Base case. For \(n=1\),

\(\frac d{dx}(x\cos x)=\cos x-x\sin x=(-1)^1(x\sin x-(2\cdot1-1)\cos x),\)

so the result is true for \(n=1\).

Inductive step. Assume the result is true for \(n=k\):

\(\frac{d^{2k-1}}{dx^{2k-1}}(x\cos x)=(-1)^k(x\sin x-(2k-1)\cos x).\)

Differentiate once:

\(\frac{d^{2k}}{dx^{2k}}(x\cos x)=(-1)^k(x\cos x+2k\sin x).\)

Differentiate again:

\(\frac{d^{2k+1}}{dx^{2k+1}}(x\cos x)=(-1)^k(-x\sin x+\cos x+2k\cos x).\)

Thus

\(\frac{d^{2k+1}}{dx^{2k+1}}(x\cos x)=(-1)^{k+1}(x\sin x-(2k+1)\cos x).\)

This is the required result for \(n=k+1\). Therefore, by induction, the identity is true for all positive integers \(n\).

Mark scheme

Answer: (a) \(u_n<5\) for all positive integers \(n\).

(b) \(u_{n+1}>u_n\) for \(n\ge1\).

(a)

Base case: \(u_1<5\), which is given.

Assume that \(u_k<5\) for some positive integer \(k\). Then

\(5-u_{k+1}=5-\frac{6u_k+5}{u_k+2}=\frac{5(u_k+2)-(6u_k+5)}{u_k+2}.\)

So

\(5-u_{k+1}=\frac{5-u_k}{u_k+2}.\)

Since \(u_k<5\), the numerator is positive. Since the sequence is positive, \(u_k+2>0\). Hence \(5-u_{k+1}>0\), so \(u_{k+1}<5\).

Therefore, by induction, \(u_n<5\) for all positive integers \(n\).

(b)

Consider the difference:

\(u_{n+1}-u_n=\frac{6u_n+5}{u_n+2}-u_n.\)

Then

\(u_{n+1}-u_n=\frac{6u_n+5-u_n(u_n+2)}{u_n+2}=\frac{-u_n^2+4u_n+5}{u_n+2}.\)

Factorise the numerator:

\(-u_n^2+4u_n+5=(5-u_n)(u_n+1).\)

So

\(u_{n+1}-u_n=\frac{(5-u_n)(u_n+1)}{u_n+2}.\)

From part (a), \(5-u_n>0\), and since \(u_n>0\), the denominator and \(u_n+1\) are positive. Hence \(u_{n+1}-u_n>0\), so \(u_{n+1}>u_n\).

Mark scheme