Mathematical Induction

Given $A = \left[ {\matrix{ 1 & 0 \cr 1 & 1 \cr } } \right]$

$\therefore$ $A \times A$ = ${A^2}$ = $\left[ {\matrix{ 1 & 0 \cr 2 & 1 \cr } } \right]$

and ${A^3}$ = ${A^2} \times A$ = $\left[ {\matrix{ 1 & 0 \cr 3 & 1 \cr } } \right]$

So we can say ${A^n}$ = $\left[ {\matrix{ 1 & 0 \cr n & 1 \cr } } \right]$

Now $nA - \left( {n - 1} \right){\rm I}$

= $\left[ {\matrix{ n & 0 \cr n & n \cr } } \right]$ - $\left[ {\matrix{ {n - 1} & 0 \cr 0 & {n - 1} \cr } } \right]$

= $\left[ {\matrix{ 1 & 0 \cr n & 1 \cr } } \right]$ = ${A^n}$

$\therefore$ ${A^n} = nA - \left( {n - 1} \right){\rm I}$

Given $S(K)$ $ = 1 + 3 + 5... + \left( {2K - 1} \right) = 3 + {K^2}$

When k = 1, S(1): 1 = 3 + 1,

L.H.S of S(k) $ \ne $ R.H.S of S(k)

So S(1) is not true.

As S(1) is not true so principle of mathematical induction can not be used.

S(K+1) = 1 + 3 + 5... + (2K - 1) + (2K + 1) = 3 + (k + 1)²

Now let S(k) is true

$\therefore$ 1 + 3 + 5 +........(2k - 1) = 3 + k²

$ \Rightarrow $ 1 + 3 + 5 +........(2k - 1) + (2k + 1) = 3 + k² + 2k +1

= 3 + (k + 1)²

$ \Rightarrow $ S(k + 1) is true.

$\therefore$ S(k) $ \Rightarrow $ S(k + 1)

Given ${a_n} = \sqrt {7 + \sqrt {7 + \sqrt {7 + .......} } } $

$\therefore$ ${a_n} = \sqrt {7 + {a_n}} $

$ \Rightarrow $ $a_n^2 = 7 + {a_n}$

$ \Rightarrow $ $a_n^2 - {a_n} - 7 = 0$

$ \Rightarrow {a_n} = {{1 \pm \sqrt {1 - 4 \times 1 \times - 7} } \over 2}$

$ \Rightarrow {a_n} = {{1 \pm \sqrt {29} } \over 2}$

As ${a_n}$ > 0,

$\therefore$ ${a_n} = {{1 + \sqrt {29} } \over 2}$ = 3.19

So ${a_n} > 3\,\,\forall \,\,n \ge 1$