Math, asked by adarshverma19, 4 months ago

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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Answers

Answered by RiyaSethi17
0

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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