31.
If $${a_n} = \sqrt {7 + \sqrt {7 + \sqrt {7 + ......} } } $$ having $$n$$ radical signs then by methods of mathematical induction which is true
A
$${a_n} > 7\,\,\forall \,\,n \geqslant 1$$
B
$${a_n} < 7\,\,\forall \,\,n \geqslant 1$$
C
$${a_n} < 4\,\,\forall \,\,n \geqslant 1$$
D
$${a_n} < 3\,\,\forall \,\,n \geqslant 1$$
Answer :
$${a_n} < 7\,\,\forall \,\,n \geqslant 1$$
$$\eqalign{
& {a_1} = \sqrt 7 < 7.\,\,{\text{Let }}{a_m} < 7 \cr
& {\text{Then }}{a_{m + 1}} = \sqrt {7 + {a_m}} \cr
& \Rightarrow \,\,{a^2}_{m + 1} = 7 + {a_m} < 7 + 7 < 14. \cr
& \Rightarrow \,\,{a_{m + 1}} < \sqrt {14} < 7; \cr} $$
So by the principle of mathematical induction $${a_n} < 7\,\,\forall \,\,n.$$
32.
For all $$n \in N,{41^n} - {14^n}$$ is a multiple of
A
26
B
27
C
25
D
None of these
Answer :
27
Let $$P\left( n \right)$$ be the statement given by $$P\left( n \right):{41^n} - {14^n}$$ is a multiple of 27
For $$n = 1,$$
i.e., $$P\left( 1 \right) = {41^1} - {14^1} = 27 = 1 \times 27,$$
which is a multiple of 27.
$$\therefore P\left( 1 \right)$$ is true.
Let $$P\left( k \right)$$ be true, i.e., $${41^k} - {14^k} = 27\lambda \,\,\,\,\,.....\left( {\text{i}} \right)$$
For $$n = k + 1,$$
$$\eqalign{
& {41^{k + 1}} - {14^{k + 1}} = {41^k}41 - {14^k}14 \cr
& = \left( {27\lambda + {{14}^k}} \right)41 - {14^k}14\,\,\,\left[ {{\text{using}}\left( {\text{i}} \right)} \right] \cr
& = \left( {27\lambda \times 41} \right) + \left( {{{14}^k} \times 41} \right) - \left( {{{14}^k} \times 14} \right) \cr
& = \left( {27\lambda \times 41} \right) + {14^k}\left( {41 - 14} \right) \cr
& = \left( {27\lambda \times 41} \right) + \left( {{{14}^k} \times 27} \right) \cr
& = 27\left( {41\lambda + {{14}^k}} \right), \cr} $$
which is a multiple of 27.
Therefore, $$P\left( {k + 1} \right)$$ is true when $$P\left( {k} \right)$$ is true.
Hence, from the principle of mathematical induction, the statement is true for all natural numbers $$n.$$