Statement 1: If A = 0a b0 then An = 0an bn0 for all n ∈ NStatement 2: If P = 13 -1-4 then Pn = n1+2n 1-2n-4n for all n ∈ NThen,

Statement 1: A =

[_{0}^{a}_{b}^{0}]

To prove B(n) =

A^{n}

[_{0}^{a^{n}}_{b^{n}}^{0}]

By mathematical induction, for n = 1, we get,

A^{1}

[_{0}^{a}_{b}^{0}]

Now let n = k, then

A^{k}

[_{0}^{a^{k}}_{b^{k}}^{0}]

Similarly, for n = k+1, then

A^{k + 1}

[_{0}^{a^{k + 1}}_{b^{k + 1}}^{0}]

should be the result
Now, let n = k+1

A^{k + 1}

A^{k} . A

[_{0}^{a^{k}}_{b^{k}}^{0}]

[_{0}^{a}_{b}^{0}]

[_{0 \times a + b^{k} \times 0}^{a^{k} \times a + 0 \times 0}_{{0 \times 0 + b}^{k} \times b}^{a^{k} \times 0 + 0 \times b}]

[_{0}^{a^{k + 1}}_{b^{k + 1}}^{0}]

Statement 2: Given, P =

[_{1}^{3}_{- 1}^{- 4}]

To prove

P^{n}

[_{n}^{1 + 2 n}_{1 - 2 n}^{- 4 n}]

By the principle of mathematical induction,
Let n = 1, then,

P^{1}

[_{1}^{1 + 2 (1)}_{1 - 2 (1)}^{- 4 (1)}]

[_{1}^{3}_{- 1}^{- 4}]

It is true for n=1
Let n=k, then,

P^{k}

[_{k}^{1 + 2 k}_{1 - 2 k}^{- 4 k}]

For n=k+1, we get,

P^{k + 1}

[_{k + 1}^{1 + 2 (k + 1)}_{1 - 2 (k + 1)}^{- 4 (k + 1)}]

………….(1)
We know,

P^{k + 1}

P^{k} . P^{1}

[_{k}^{1 + 2 k}_{1 - 2 k}^{- 4 k}]

[_{1}^{3}_{- 1}^{- 4}]

[_{3 k + 1 (1 - 2 k)}^{3 + 6 k - 4 k}_{- 4 k - 1 (1 - 2 k)}^{- 4 (1 + 2 k) + 4 k}]

[_{1 + k}^{3 + 2 k}_{- 1 - 2 k}^{- 4 - 4 k}]

[_{k + 1}^{1 + 2 (k + 1)}_{1 - 2 (k + 1)}^{- 4 (k + 1)}]

………………..(2)
From (1) and (2), we conclude result is true for n = k+1

Statement 1: If A = $[_{0}^{a}_{b}^{0}]$ then $A^{n}$ = $[_{0}^{a^{n}}_{b^{n}}^{0}]$ for all n $\in$ N
Statement 2: If P = $[_{1}^{3}_{- 1}^{- 4}]$ then $P^{n}$ = $[_{n}^{1 + 2 n}_{1 - 2 n}^{- 4 n}]$ for all n $\in$ N
Then,