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Solve : 52n + 2 – 24n – 25 is divisible by 576 for all n ϵ N
CBSE | Class 11 | Exercise 12.2 | Mathematical Induction | Maths | RD Sharma
Solve : 52n + 2 – 24n – 25 is divisible by 576 for all n ϵ N

Let ,

    \[P\text{ }\left( n \right):\text{ }{{5}^{2n\text{ }+\text{ }2}}~\text{ }24n\text{ }\text{ }25\text{ }is\text{ }divisible\text{ }by\text{ }576\]

Let us check for

    \[n\text{ }=\text{ }1,\]

    \[P\text{ }\left( 1 \right):\text{ }{{5}^{2.1+2}}~\text{ }24.1\text{ }\text{ }25\]

    \[:\text{ }625\text{ }\text{ }49\]

    \[:\text{ }576\]

P (n) is true for n = 1. Where, P (n) is divisible by 576

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1)

is true.

P (k): 52k + 2 – 24k – 25 is divisible by 576

    \[:\text{ }{{5}^{2k\text{ }+\text{ }2}}~\text{ }24k\text{ }\text{ }25\text{ }=\text{ }576\lambda \text{ }\ldots .\text{ }\left( i \right)\]

We have to prove,

52k + 4 – 24(k + 1) – 25 is divisible by 576

    \[{{5}^{\left( 2k\text{ }+\text{ }2 \right)\text{ }+\text{ }2}}~\text{ }24\left( k\text{ }+\text{ }1 \right)\text{ }\text{ }25\text{ }=\text{ }576\mu \]

So,

    \[=~{{5}^{\left( 2k\text{ }+\text{ }2 \right)\text{ }+\text{ }2}}~\text{ }24\left( k\text{ }+\text{ }1 \right)\text{ }\text{ }25\]

    \[=\text{ }{{5}^{(2k\text{ }+\text{ }2)}}{{.5}^{2}}~\text{ }24k\text{ }\text{ }24\text{ }25\]

    \[=\text{ }\left( 576\lambda ~+\text{ }24k\text{ }+\text{ }25 \right)25\text{ }\text{ }24k\text{ }49\]

by using equation (i)

    \[=\text{ }25.\text{ }576\lambda \text{ }+\text{ }576k\text{ }+\text{ }576\]

    \[=\text{ }576\left( 25\lambda \text{ }+\text{ }k\text{ }+\text{ }1 \right)\]

    \[=\text{ }576\mu \]

P (n) is true for

    \[n\text{ }=\text{ }k\text{ }+\text{ }1\]

Hence, P (n) is true for all n ∈ N.

i

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