A) 2
B) 3
C) 7
D) 1
Correct Answer: D
Solution :
| [d] Let P (n) be the statement given by |
| \[P(n):{{3}^{2n}}\] when divided by 8, the remainder is 1. or \[P(n):{{3}^{2n}}=8\lambda +1\] for some \[\lambda \in N\] |
| For \[n=1,\,\,\,P(1):{{3}^{2}}=(8\times 1)+1=8\lambda +1,\] where \[\lambda =1\] |
| \[\therefore \,\,\,P(1)\] is true. |
| Let P (k) be true. |
| Then \[{{3}^{2k}}=8\lambda +1\] for some \[\lambda \in N\] ? (i) |
| We shall now show that \[P(k+1)\] is true, for which we have to show that \[{{3}^{2(k+1)}}\] when divided by 8, the remainder is 1. |
| Now \[{{3}^{2(k+1)}}={{3}^{2k}}{{.3}^{2}}\] |
| \[=(8\lambda +1)\times 9\] [Using (i)] |
| \[=72\lambda +9=72\lambda +8+1=8(9\lambda +1)+1\] |
| \[=8\mu +1,\] where \[\mu =9\lambda +1\in N\] |
| \[\Rightarrow P(k+1)\] is true. |
| Thus, \[P(k+1)\] is true, whenever P(k) is true. Hence, by the principle of mathematical induction \[P(n)\] is true for all \[n\in N\]. |
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