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By MI
n=1
21x18^2 +36x7^3
=19152
=19x1008
is divisible by 19
Assume p(k) is true
ie
21x18^(2k) +36x7^(3k) =19m
for n=k+1
21x18^(2k+2) +36x7^(3k+3)
=324x[21x18^(2k)] + 343x[36x7^(3k)]
=324x[21x18^(2k)] + 324x[36x7^(3k)] + 19x[36x7^(3k)]
=324x[21x18^(2k+2) +36x7^(3k+3)] +19x[36x7^(3k)]
=324x19m +19x[36x7^(3k)]
=19(324m + 36x7^(3k) )
is divisible by 19
p(k+1) is true
By principle of MI , it is true for all +ve integer n
n=1
21x18^2 +36x7^3
=19152
=19x1008
is divisible by 19
Assume p(k) is true
ie
21x18^(2k) +36x7^(3k) =19m
for n=k+1
21x18^(2k+2) +36x7^(3k+3)
=324x[21x18^(2k)] + 343x[36x7^(3k)]
=324x[21x18^(2k)] + 324x[36x7^(3k)] + 19x[36x7^(3k)]
=324x[21x18^(2k+2) +36x7^(3k+3)] +19x[36x7^(3k)]
=324x19m +19x[36x7^(3k)]
=19(324m + 36x7^(3k) )
is divisible by 19
p(k+1) is true
By principle of MI , it is true for all +ve integer n
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感谢感谢,还有两题有可能的话还请大神解答
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