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How many positive integers n are there such that 3sns 100 and x +x+1 is divisible by x +x+1? |
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Answer» hey mate!! here is your answer Step-by-step explanation: Let A n
=x(x n−1 −na n−1 )+a n (n−1) For n = 2, A 2
=x(x−2a)+a 2
=x 2 −2axa 2 =(x−a) 2 . Thus A 2
is divisible by (x−a) 2 . Now assume that A m
is divisible by (x−a) 2 for m ≥ 2, that is, assume A m
=x(x m−1 −ma m−1 )+a m (m−1) =(x−a) 2 F (x) so that x m =mxa m−1 −a m (m−1)+(x−a) 2 f(x) We then have A m−1
=x[x m −(m+1)a m ]+a m+1 m =x.x m −(m+1)xa m +ma m,+1
=x[mxa m−1 −a m (m−1)+(x−a) 2 f(x)] −(m+1)xa m +ma m+1 by =ma m−1 [(x 2 −2xa+a 2 )+x(x−a) 2 f(x)] =ma m−1 ((x−a) 2 +x(x−a) 2 f(x) =(x−a) 2 [ma m−1 +xf(x)], After simplification. This shows that A m+1
is divisible by (x−a) 2 . Hence by INDUCTION, A n
is divisible by (x−a) 2 for all positive INTEGERS n > 1. |
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