InterviewSolution
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InterviewSolution
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Using principle of mathematical induction, prove that 7^(4^(n)) -1 is divisible by 2^(2n+3) for any naturalnumber n. |
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Answer» <P> Solution :Let `P(n) = 7^(4^(th)) - 1` be divisible by `2^(2n+3)``P(1) = 7^(4) - 1= (7^(2) - 1) (7^(2) + 1)` ` = 48 xx 50= 32 xx 75` `= 2^(5) xx 75`. whichis divisibleby `2^(2xx1+3)` Let usassume that the result is truefor n` = k`. i.e, `7^(4^(k)) - I` is divisible by `2^(2k+3)`but not by `2^(2k+4)`. `rArr 7^(4^(k)) - 1 = 2^(2k+3)`m , wherem issome oddnatural number Now, `7^(4^(k+1)) - 1 = (7^(4^(k)))^(4-1)` `= ((2^(2k+3)m+1)^(2) +1)((2^(2k+3)m+1)^(2)-1)` `= ((2^(2k+3)m+1)^(2) +1)(2^(2k+3)+2)(2^(2k+3)m)` `= (2^(4k+6)m^(2) + 2^(2k+4) m + 2)(2^(2k+3)m + 2) (2^(2k+3)m)` ltbr `= 2^(2k-5)(2^(4k+5)m^(2) + 2^(2k+3) m + 1)(2^(2^(2k+2)m +1) (m),` Which is divisivble by `2^(2k+5)` Thus, `P(k + 1)` is true whenever `P(k)`is true. So, by the PRINCIPLE of MATHEMATICAL induction, `P(n)` is true for any natural number n. |
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