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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