For each, $n \in N$, $10^{2n - 1} + 1$ is divisible by

Question

Dr. Abubacker · Accepted Answer

(a)
Let $P(n) = 10^{2n - 1} + 1$
$P(1) = 10 + 1 = 11$
Let $P(k) \equiv 10^{2k - 1} + 1 = 11I\$ is true
Now, $P(k + 1) = 10^{2k + 1} + 1$
$= (11I - 1)100 + 1$
$= 1100I - 99 = 11I_{1}$
So, $P(k + 1)$ is true.

Q. For each, $n \in N$, $10^{2n - 1} + 1$ is divisible by

Explanation

Reviewed by Dr. Abubacker

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