1. (a) The problem states two conditions for an arithmetic progression (AP): the sum of the first 6 terms $S_6 = 72$ and the 8th term $a_8 = 16$. (i) Recall the formula for the $n^{th}$ term of an AP: $a_n = a_1 + (n-1)d$ where $a_1$ is the first term and $d$ is the common difference. For the 8th term: $$a_8 = a_1 + 7d = 16$$ (ii) The sum of the first $n$ terms is given by: $$S_n = \frac{n}{2} [2a_1 + (n-1)d]$$ For $n=6$: $$72 = \frac{6}{2} [2a_1 + 5d] = 3(2a_1 + 5d)$$ Divide both sides by 3: $$24 = 2a_1 + 5d$$ Now we have two equations: $$\begin{cases} a_1 + 7d = 16 \\ 2a_1 + 5d = 24 \end{cases}$$ Multiply the first equation by 2: $$2a_1 + 14d = 32$$ Subtract the second equation: $$(2a_1 + 14d) - (2a_1 + 5d) = 32 - 24 \\ 9d = 8 \\ d = \frac{8}{9}$$ Substitute $d$ back into $a_1 + 7d = 16$: $$a_1 + 7\times \frac{8}{9} = 16 \\ a_1 + \frac{56}{9} = 16 \\ a_1 = 16 - \frac{56}{9} = \frac{144}{9} - \frac{56}{9} = \frac{88}{9}$$ (b) The sum of the first 7 terms of two APs $A_k$ and $B_k$ are equal: $$\sum_{k=1}^7 A_k = \sum_{k=1}^7 B_k$$ No further information is given, so the problem likely expects recognition of this equality or uses it in a further problem. (c) (i) Apply elementary row operations to transform matrix $A$ to upper triangular form: $$A = \begin{bmatrix}2 & 6 & 1 \\ 5 & 2 & 3 \\ -2 & 1 & 4\end{bmatrix}$$ Step 1: Use $R_1$ to eliminate below: $$R_2 \to R_2 - \frac{5}{2}R_1 = \begin{bmatrix}5 - 5 & 2 - 15 & 3 - \frac{5}{2}\times 1\end{bmatrix} = \begin{bmatrix}0 & -13 & 0.5\end{bmatrix}$$ $$R_3 \to R_3 + R_1 = \begin{bmatrix}-2 + 2 & 1 + 6 & 4 + 1\end{bmatrix} = \begin{bmatrix}0 & 7 & 5\end{bmatrix}$$ Matrix now: $$\begin{bmatrix}2 & 6 & 1 \\ 0 & -13 & 0.5 \\ 0 & 7 & 5 \end{bmatrix}$$ Step 2: Use $R_2$ to eliminate $R_3$ second element: $$R_3 \to R_3 - \frac{7}{-13} R_2 = R_3 + \frac{7}{13} R_2 = \begin{bmatrix}0 & 0 & 5 + \frac{7}{13}\times 0.5 \end{bmatrix} = \begin{bmatrix}0 & 0 & 5 + \frac{3.5}{13}\end{bmatrix} = \begin{bmatrix}0 & 0 & \frac{65}{13} + \frac{3.5}{13} = \frac{68.5}{13} \approx 5.27\end{bmatrix}$$ (ii) Find the reduced row echelon form (RREF) of matrix $B$: $$B = \begin{bmatrix}0 & 2 & 5 \\ 1 & 0 & 1 \\ 0 & 1 & 3\end{bmatrix}$$ Step 1: Swap $R_1 \leftrightarrow R_2$ to get leading 1: $$\begin{bmatrix}1 & 0 & 1 \\ 0 & 2 & 5 \\ 0 & 1 & 3\end{bmatrix}$$ Step 2: $R_2 \to \frac{R_2}{2} = \begin{bmatrix}0 & 1 & \frac{5}{2} \end{bmatrix}$ Step 3: Eliminate second column below and above: $$R_3 \to R_3 - R_2 = \begin{bmatrix}0 & 0 & 3 - \frac{5}{2} = \frac{1}{2}\end{bmatrix}$$ $$R_1 \to R_1 - 0 \times R_2 = R_1$$ Step 4: Normalize $R_3$: $$R_3 \to 2R_3 = \begin{bmatrix}0 & 0 & 1 \end{bmatrix}$$ Step 5: Eliminate third column above: $$R_2 \to R_2 - \frac{5}{2} R_3 = \begin{bmatrix}0 & 1 & 0\end{bmatrix}$$ $$R_1 \to R_1 - 1 \times R_3 = \begin{bmatrix}1 & 0 & 0\end{bmatrix}$$ RREF is: $$\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ (iii) Determine rank of $C$: $$C = \begin{bmatrix}1 & 2 & 3 \\ 2 & 5 & 6 \\ 3 & 8 & 9\end{bmatrix}$$ Perform row operations: $$R_2 \to R_2 - 2R_1 = \begin{bmatrix}0 & 1 & 0\end{bmatrix}$$ $$R_3 \to R_3 - 3R_1 = \begin{bmatrix}0 & 2 & 0\end{bmatrix}$$ Then $R_3 \to R_3 - 2R_2 = \begin{bmatrix}0 & 0 & 0\end{bmatrix}$ Rank is the number of nonzero rows: 2. 2. (a) (i) Expand $$\left(2x - \frac{1}{2x^2}\right)^5$$ Use binomial theorem: $$\sum_{k=0}^5 \binom{5}{k} (2x)^{5-k} \left(-\frac{1}{2x^2}\right)^k$$ Simplify terms: Each term: $$\binom{5}{k} 2^{5-k} x^{5-k} (-1)^k 2^{-k} x^{-2k} = \binom{5}{k} (-1)^k 2^{5-k-k} x^{5-k-2k} = \binom{5}{k} (-1)^k 2^{5-2k} x^{5-3k}$$ Write terms explicitly: For $k=0: 1 \times 2^{5} x^{5} = 32x^5$ $k=1: -5 \times 2^{3} x^{2} = -5 \times 8 x^{2} = -40x^2$ $k=2: 10 \times 2^{1} x^{-1} = 20x^{-1}$ $k=3: -10 \times 2^{-1} x^{-4} = -5x^{-4}$ $k=4: 5 \times 2^{-3} x^{-7} = \frac{5}{8} x^{-7}$ $k=5: -1 \times 2^{-5} x^{-10} = -\frac{1}{32} x^{-10}$ Expanded form: $$32x^5 - 40x^2 + 20x^{-1} - 5x^{-4} + \frac{5}{8} x^{-7} - \frac{1}{32} x^{-10}$$ (b) Prove by induction: $$\sum_{k=1}^n 2^k = 2^{n+1} - 2$$ *Base case* $n=1$: $$2^1 = 2 \quad \text{and} \quad 2^{1+1} - 2 = 4 - 2 = 2$$ True. *Inductive step* Assume true for $n=m$: $$\sum_{k=1}^m 2^k = 2^{m+1} - 2$$ Show true for $n=m+1$: $$\sum_{k=1}^{m+1} 2^k = \sum_{k=1}^m 2^k + 2^{m+1} = (2^{m+1} - 2) + 2^{m+1} = 2 \times 2^{m+1} - 2 = 2^{m+2} -2$$ Holds true. (c) (i) Find the term independent of $x$ in $$\left(x^2 + \frac{1}{x}\right)^5$$ General term: $$T_{k+1} = \binom{5}{k} (x^2)^{5-k} \left(\frac{1}{x}\right)^k = \binom{5}{k} x^{2(5-k)} x^{-k} = \binom{5}{k} x^{10 - 2k - k} = \binom{5}{k} x^{10 - 3k}$$ Term is independent of $x$ when exponent is 0: $$10 - 3k = 0 \implies k = \frac{10}{3}$$ Since $k$ must be integer between 0 and 5, no integer $k$ makes term independent of $x$. Hence there is no term independent of $x$. (ii) Prove by induction: $$1 + 2 + 2^2 + \cdots + 2^{n-1} = 2^n - 1$$ *Base case* $n=1$: $$2^{0} = 1 \quad \text{and} \quad 2^1 - 1 = 1$$ True. *Inductive step* Assume true for $n=m$: $$\sum_{k=0}^{m-1} 2^k = 2^m - 1$$ Show true for $n=m+1$: $$\sum_{k=0}^m 2^k = (2^m - 1) + 2^m = 2 \times 2^m - 1 = 2^{m+1} - 1$$ True. Final answers: 1.(a)(i) $a_1=\frac{88}{9}$ 1.(a)(ii) $d=\frac{8}{9}$ 1.(c)(i) Upper triangular matrix: $$\begin{bmatrix}2 & 6 & 1 \\ 0 & -13 & 0.5 \\ 0 & 0 & 5.27\end{bmatrix}$$ 1.(c)(ii) RREF of $B$ is identity matrix: $$\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$$ 1.(c)(iii) rank of $C$ is 2 2.(a)(i) Expanded: $$32x^5 - 40x^2 + 20x^{-1} - 5x^{-4} + \frac{5}{8}x^{-7} - \frac{1}{32} x^{-10}$$ 2.(b) and 2.(c)(ii) both proofs by induction shown. 2.(c)(i) no term independent of $x$ in the expansion.