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 \[n.\]