Đáp án:

$I = {100^{{2^n}}} - 1$

Giải thích các bước giải:

 Ta có:

$\begin{array}{l}
I = 101.10001.100000001....100...0001\\
 = \left( {100 + 1} \right)\left( {{{100}^2} + 1} \right)\left( {{{100}^4} + 1} \right)...\left( {{{100}^{{2^{n - 1}}}} + 1} \right)
\end{array}$

$ \Rightarrow I\left( {100 - 1} \right) = \left( {100 - 1} \right)\left( {100 + 1} \right)\left( {{{100}^2} + 1} \right)\left( {{{100}^4} + 1} \right)...\left( {{{100}^{{2^{n - 1}}}} + 1} \right)$

Áp dụng HĐT: ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right),\forall a,b$ ta có:

$\begin{array}{l}
I\left( {100 - 1} \right) = \left( {{{100}^2} - 1} \right)\left( {{{100}^2} + 1} \right)\left( {{{100}^4} + 1} \right)...\left( {{{100}^{{2^{n - 1}}}} + 1} \right)\\
 = \left( {{{100}^4} - 1} \right)\left( {{{100}^4} + 1} \right)...\left( {{{100}^{{2^{n - 1}}}} + 1} \right)\\
 = \left( {{{100}^{{2^{n - 1}}}} - 1} \right)\left( {{{100}^{{2^{n - 1}}}} + 1} \right)\\
 = {\left( {{{100}^{{2^{n - 1}}}}} \right)^2} - 1\\
 = {100^{{2^n}}} - 1
\end{array}$

Vậy $I = {100^{{2^n}}} - 1$