We can compose the given assertion as

$P(n):{{10}^{2}}n-1+1$is distinct by \[11\]

On the off chance that \[n\text{ }=\text{ }1\]we get

$P(1)={{10}^{2}}.1-1+1=11$, which is distinct by \[11\]

Which is valid.

Think about \[P\text{ }\left( k \right)\]be valid for some sure whole number \[k\]

${{10}^{(2k-1)}}+1$is separable by \[11\]

${{10}^{(2k-1)}}+1=11m$, where \[m\in N\ldots \left( 1 \right)\]

Presently let us demonstrate that \[P\text{ }\left( k\text{ }+\text{ }1 \right)\]is valid.

Here

${{10}^{2}}(k+1)-1+1$

We can compose it as

$\begin{align}

& ={{10}^{2}}k+2-1+1 \\

& ={{10}^{2}}k+1+1 \\

\end{align}$

By expansion and deduction of \[1\]

$={{10}^{2}}(102k-1+1-1)+1$

We get

$={{10}^{2}}(102k-1+1)-102+1$

Utilizing condition \[1\] we get

$={{10}^{2}}.11m-100+1$

\[=\text{ }100\text{ }\times \text{ }11m\text{ }\text{ }99\]

Taking out the normal terms

\[=\text{ }11\text{ }\left( 100m\text{ }\text{ }9 \right)\]

\[=\text{ }11\text{ }r,\]where \[r\text{ }=\text{ }\left( 100m\text{ }\text{ }9 \right)\] is some regular number

${{10}^{2}}(k+1)-1+1$is separable by \[11\]

\[P\text{ }\left( k\text{ }+\text{ }1 \right)\]is valid at whatever point \[P\text{ }\left( k \right)\]is valid.

Thus, by the rule of numerical acceptance, explanation \[P\text{ }\left( n \right)\]is valid for all regular numbers for example \[n.\]