PROVE BY M.I (41)n-(14) is multiple of 27

The question is wrong its 41n-41n

p(k)=41k-14k 

=27m

41k=27m+14k

p(k+1)=41k+1-14k+1

=41k.41-14k.14

=(27m+14k)41-14k.14

=(27m)41+(14k)41-14k.14

=(27m)41+14k(41-14)

=(27m)41+14k(27)

=27(41m+14k)

=27q, where q=(41m+14k)

  • 25

Ya question is wrong is 41n-14n

p(k)=41k-14k

=27m

41k=27m+14k

p(k+1)=41k+1-14k+1

=41k.41-14k.14

=(27m+14k)41-14k.14

=(27m)41+(14k)41-14k.14

=(27m)41+14k(41-14)

=(27m)41+14k(27)

=27(41m+14k)

=27q, where q=(41m+14k)

  • -1
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