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Question

11n+2+122n+1 is divisible by 133 for all nϵN.

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Solution

Let P(n):11n+2+122n+1 is divisible by 133

For n = 1

=113+123

=1331+1728

=3059

It is divisible of 133

P(n) is true for n = 1

Let P(n) is true for n = k, so

11k+2+122k+1 is divisible by 133

11k+2+122k+1=133λ ........(1)

We have to show that,

11k+3+122k+3 is divisible by 133

Now,

=11k+2,11+122k+1.122

=(133λ122k+1)11+122k+1.144

=11.133λ11.122k+1+144.122k+1

=11.133λ133.122k+1

=133(11λ+122k+1)

=133μ

P(n) is true for n = k + 1

P(n) is true for all nϵN by PMI.


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