Question

7²ⁿ + 2³ⁿ⁻³. 3ⁿ⁻¹ is divisible by 25 for all n ∈ N.

 

Answer 1

Let P(n) be the given statement.
Now,

\[P(n): 7^{2n} + 2^{3n - 3} . 3^{n - 1} \text{ is divisible by } 25 . \]
\[\text{ Step1} : \]
\[P(1): 7^2 + 2^{3 - 3} . 3^{1 - 1} = 49 + 1 = 50 \]
\[\text{ It is divisible by 25}  . \]
\[\text{ Thus, P(1) is true} \]
\[\text{ Step2: Let } P\left( m \right) \text{ be true . } \]
\[\text{ Now } , \]
\[ 7^{2m} + 2^{3m - 3} . 3^{m - 1} \text{ is divisible by}  25 . \]
\[Suppose: \]
\[ 7^{2m} + 2^{3m - 3} . 3^{m - 1} = 25\lambda . . . (1)\]
\[\text{ We have to show that } P\left( m + 1 \right) \text{ is true whenever P(m) is true}  . \]
\[\text{ Now, } \]
\[P\left( m + 1 \right) = 7^{2m + 2} + 2^{3m} . 3^m \]
\[ = 7^{2m + 2} + 7^2 . 2^{3m - 3} . 3^{m - 1} - 7^2 . 2^{3m - 3} . 3^{m - 1} + 2^{3m} . 3^m \]
\[ = 7^2 \left( 7^{2m} + 2^{3m - 3} . 3^{m - 1} \right) + 2^{3m} . 3^m \left( 1 - \frac{49}{24} \right)\]
\[ = 7^2 \times 25\lambda - 2^{3m} . 3^m \times \frac{25}{2^3 . 3^1} \left[ \text{ Using } (1) \right]\]
\[ = 25\left( 49\lambda - 2^{3m - 3} . 3^{m - 1} \right) \]
\[\text{ It is divisible by } 25 . \]
\[\text{ Thus, } P\left( m + 1 \right) \text{ is true .}  \]
\[\text{ By the principle of mathematical induction, P(n) is true for all n } \in N .\]