A) 26
B) 27
C) 25
D) None of these
Correct Answer: B
Solution :
[b] Let P(n) be the statement given by \[P(n):{{41}^{n}}-{{14}^{n}}\] is a multiple of 27 |
For \[n=1,\] |
i.e., \[P(1)={{41}^{1}}-{{14}^{1}}=27=1\times 27,\] |
which is a multiple of 27. |
\[\therefore \,\,\,\,P(1)\] is true. |
Let \[P(k)\] be true, i.e, \[{{41}^{k}}-{{14}^{k}}=27\lambda \] (i) |
For \[n=k+1,\] |
\[{{41}^{k+1}}-{{14}^{k+1}}={{41}^{k}}41-{{14}^{k}}14\] |
\[=(27\lambda +{{14}^{k}})41-{{14}^{k}}14\] [using (i)] |
\[=(27\lambda \times 41)+({{14}^{k}}\times 41)-({{14}^{k}}\times 14)\] |
\[=(27\lambda \times 41)+{{14}^{k}}(41-14)=(27\lambda \times 41)+({{14}^{k}}\times 27)\]\[=27(41\lambda +{{14}^{k}}),\] |
which is a multiple of 27. |
Therefore, P(K+1) is true when P (K) is true hence. From the principle of mathematical induction, the statement is true for all natural numbers n. |
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