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Home » Prove the following by using the principle of mathematical induction for all n ∈ N: 32n + 2 – 8n – 9 is divisible by 8
Prove the following by using the principle of mathematical induction for all n ∈ N: 32n + 2 – 8n – 9 is divisible by 8
CBSE | Class 11 | Exercise 4.1 | Maths | NCERT | Principle of Mathematical Induction
Prove the following by using the principle of mathematical induction for all n ∈ N: 32n + 2 – 8n – 9 is divisible by 8

We can compose the given assertion as

P(n):{{3}^{2}}n+2-8n-9is separable by

    \[8\]

In the event that

    \[n\text{ }=\text{ }1\]

we get

P(1)={{3}^{(2\text{ }\!\!\times\!\!\text{ }1)}}+2-8\text{ }\!\!\times\!\!\text{ }1-9=64, which is distinguishable by

    \[8\]

Which is valid.

Think about

    \[P\text{ }\left( k \right)\]

be valid for some sure number

    \[k\]

{{3}^{2}}k+2-8k-9is distinct by

    \[8\]

{{3}^{2}}k+2-8k-9=8m,wherem?N\text{ n}(1)

Presently let us demonstrate that

    \[P\text{ }\left( k\text{ }+\text{ }1 \right)\]

is valid.

Here

{{3}^{(2(}}k+1)+2)-8(k+1)-9

We can compose it as

={{3}^{(2k+2)}}{{.3}^{2}}-8k-8-9

By adding and taking away

    \[8k\text{ }and\text{ }9\]

we get

={{3}^{2}}({{3}^{(2k)}}+2-8k-9+8k+9)-8k-17

On additional improvement

={{3}^{2}}({{3}^{(2k)}}+2-8k-9)+{{3}^{2}}(8k+9)-8k-17

 

From condition

    \[\left( 1 \right)\]

we get

    \[=\text{ }9.\text{ }8m\text{ }+\text{ }9\text{ }\left( 8k\text{ }+\text{ }9 \right)\text{ }\text{ }8k\text{ }\text{ }17\]

By duplicating the terms

    \[=\text{ }9.\text{ }8m\text{ }+\text{ }72k\text{ }+\text{ }81\text{ }\text{ }8k\text{ }\text{ }17\]

So we get

    \[=\text{ }9.\text{ }8m\text{ }+\text{ }64k\text{ }+\text{ }64\]

By taking out the normal terms

    \[=\text{ }8\text{ }\left( 9m\text{ }+\text{ }8k\text{ }+\text{ }8 \right)\]

    \[=\text{ }8r\]

, where

    \[r\text{ }=\text{ }\left( 9m\text{ }+\text{ }8k\text{ }+\text{ }8 \right)\]

is a characteristic number

So {{3}^{(2(}}k+1)+2)-8(k+1)-9 is distinct by

    \[8\]

    \[P\text{ }\left( k\text{ }+\text{ }1 \right)\]

is valid at whatever point

    \[P\text{ }\left( k \right)\]

is valid.

 

Hence, by the rule of numerical acceptance, articulation

    \[P\text{ }\left( n \right)\]

is valid for all regular numbers for example

    \[n.\]

i

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