A) 7
B) 3
C) 4
D) 5
Correct Answer: C
Solution :
| [c] Let \[P(n):{{7}^{n}}-{{3}^{n}}\] is divisible by 4. |
| For \[n=1\], |
| \[P(1):{{7}^{1}}-{{3}^{1}}=4,\] which is divisible by 4. Thus, P (n) is true for n = 1. |
| Let P (k) be true for some natural number k, |
| i.e. \[P(k):{{7}^{k}}-{{3}^{k}}\] is divisible by 4. |
| We can write \[{{7}^{k}}-{{3}^{k}}=4d,\] where \[d\in N\] ? (i) |
| Now, we wish to prove that \[P(k+1)\] is true whenever P(k) is true, i.e., \[{{7}^{k+1}}-{{3}^{k+1}}\] is divisible by 4. |
| Now, \[{{7}^{(k+1)}}-{{3}^{(k+1)}}\] |
| \[={{7}^{(k+1)}}-{{7.3}^{k}}+{{7.3}^{k}}-{{3}^{(k+1)}}\] |
| \[=7({{7}^{k}}-{{3}^{k}})+(7-3){{3}^{k}}=7(4d)+{{4.3}^{k}}\] [Using (i)] |
| \[=4(7d+{{3}^{k}}),\] Which is divisible by 4. |
| Thus, \[P(k+1)\] is true whenever \[P(k)\] is true. Therefore, by the principle of mathematical induction the statement is true for every positive integer n. |
You need to login to perform this action.
You will be redirected in
3 sec
