1
Basis Step ($n = 1$):
$$\text{LHS} = 2(1) – 1 = 1, \quad \text{RHS} = 1^2 = 1$$
Since $\text{LHS} = \text{RHS}$, the statement holds for $n = 1$.
2
Assumption Step ($n = k$):
Assume true for $n = k$:
$$\sum_{r=1}^{k}(2r-1) = k^2$$
3
Inductive Step ($n = k + 1$):
Split the series into the first $k$ terms and the $(k+1)$-th term:
$$\sum_{r=1}^{k+1}(2r-1) = \sum_{r=1}^{k}(2r-1) + [2(k+1) – 1]$$
Substitute the assumption $k^2$:
$$= k^2 + 2k + 1 = (k+1)^2$$
4
Conclusion Step:
If true for $n = k$, then it is true for $n = k+1$. Since it is true for $n = 1$, it is true for all $n \in \mathbb{Z}^+$ by mathematical induction.
Final Answer: $\sum_{r=1}^{n}(2r-1) = n^2$ is proven for all $n \in \mathbb{Z}^+$.
1
Basis Step ($n = 1$):
$$f(1) = 3^{2(1)} + 11 = 9 + 11 = 20 = 4(5)$$
Divisible by $4$, so true for $n = 1$.
2
Assumption Step ($n = k$):
Assume $3^{2k} + 11 = 4m$ for some integer $m$.
3
Inductive Step ($n = k + 1$):
$$f(k+1) = 3^{2(k+1)} + 11 = 9(3^{2k}) + 11$$
Split to extract the $(3^{2k} + 11)$ term:
$$f(k+1) = (3^{2k} + 11) + 8(3^{2k}) = 4m + 8(3^{2k}) = 4\left[m + 2(3^{2k})\right]$$
Since $m + 2(3^{2k})$ is an integer, $f(k+1)$ is divisible by $4$.
4
Conclusion Step:
If $f(k)$ is divisible by $4$, then $f(k+1)$ is divisible by $4$. Since $f(1)$ is divisible by $4$, $f(n)$ is divisible by $4$ for all $n \in \mathbb{Z}^+$ by mathematical induction.
Final Answer: $3^{2n} + 11$ is proven divisible by $4$ for all $n \in \mathbb{Z}^+$.
1
Basis Step ($n = 1$):
$$\text{LHS} = \begin{pmatrix} 1 & -1 \\ 0 & 2 \end{pmatrix}^1, \quad \text{RHS} = \begin{pmatrix} 1 & 1-2^1 \\ 0 & 2^1 \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ 0 & 2 \end{pmatrix}$$
$\text{LHS} = \text{RHS}$, so true for $n = 1$.
2
Assumption Step ($n = k$):
Assume $\begin{pmatrix} 1 & -1 \\ 0 & 2 \end{pmatrix}^k = \begin{pmatrix} 1 & 1-2^k \\ 0 & 2^k \end{pmatrix}$.
3
Inductive Step ($n = k + 1$):
$$M^{k+1} = M^k \cdot M = \begin{pmatrix} 1 & 1-2^k \\ 0 & 2^k \end{pmatrix} \begin{pmatrix} 1 & -1 \\ 0 & 2 \end{pmatrix}$$
Multiply matrices:
$$= \begin{pmatrix} 1(1) + (1-2^k)(0) & 1(-1) + (1-2^k)(2) \\ 0(1) + 2^k(0) & 0(-1) + 2^k(2) \end{pmatrix} = \begin{pmatrix} 1 & 1 – 2^{k+1} \\ 0 & 2^{k+1} \end{pmatrix}$$
4
Conclusion Step:
Statement holds for $n = k+1$. True for all $n \in \mathbb{Z}^+$ by mathematical induction.
Final Answer: $\begin{pmatrix} 1 & -1 \\ 0 & 2 \end{pmatrix}^n = \begin{pmatrix} 1 & 1-2^n \\ 0 & 2^n \end{pmatrix}$ holds for all $n \in \mathbb{Z}^+$.
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