Solve the following
Question:

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

Solution:

Let P(n) be the given statement.

Now,

$P(n): 11^{n+2}+12^{2 n+1}$ is divisible by $133 .$

Step 1:

$P(1)=11^{1+2}+12^{2+1}=1331+1728=3059$

It is divisible by 133 .

Step 2 :

Let $P(m)$ be divisible by 133 .

Now,

$11^{m+2}+12^{2 m+1}$ is divisible by 133 .

Suppose :

$11^{m+2}+12^{2 m+1}=133 \lambda \quad \ldots$ (1)

We shall show that $P(m+1)$ is true whenever $P(m)$ is true.

Now,

$P(m+1)=11^{m+3}+12^{2 m+3}$

$=11^{m+2} \cdot 11+12^{2 m+1} \cdot 12^{2}+11 \cdot 12^{2 m+1}-11 \cdot 12^{2 m+1}$

$=11\left(11^{m+2}+12^{2 m+1}\right)+12^{2 m+1}(144-11)$

$=11.133 \lambda+12^{2 m+1} .133 \quad[$ From $(1)]$

$=133\left(11 \lambda+12^{2 m+1}\right)$

It is divisible by 133 .

Thus, $P(m+1)$ is true.

By the principle of $m$ athematical $i$ nduction, $P(n)$ is true for all $\mathrm{n} \in N$.

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