Official solution
I2 = \[\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{101}}} dx\]
I2 = \[\int\limits_0^1 {\left( {1 - {x^{50}}} \right){{\left( {1 - {x^{50}}} \right)}^{100}}dx} \]
I2 = \[\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}dx} - \int\limits_0^1 {{x^{50}}{{\left( {1 - {x^{50}}} \right)}^{100}}dx} \]
I2 = I1 - \[\int\limits_0^1 {x.{x^{49}}{{\left( {1 - {x^{50}}} \right)}^{100}}dx} \]
Now apply IBP
I2 = I1 -
\[\left[ {x\int\limits_0^1 {{x^{49}}{{\left( {1 - {x^{50}}} \right)}^{100}}dx} - \int {{{d\left( x \right)} \over {dx}}.\int {{{d\left( x \right)} \over {dx}}\int\limits_0^1 {{x^{49}}{{\left( {1 - {x^{50}}} \right)}^{100}}dx} } } } \right]\]
Let (1 -- x50) = t
\[ \Rightarrow \] -50x49dx = dt
I2 = I1 -
\[\left[ {x.\left( { - {1 \over {50}}} \right){{{{\left( {1 - {x^{50}}} \right)}^{101}}} \over {101}}} \right]_0^1\] - \[ - \int\limits_0^1 {\left( { - {1 \over {50}}} \right){{{{\left( {1 - {x^{50}}} \right)}^{101}}} \over {101}}dx} \]
= I1 - 0 - \[{ - {1 \over {50}}.{1 \over {101}}{I_2}}\]
I2 = I1 - \[{1 \over {5050}}{I_2}\]
\[ \Rightarrow \] I2 + \[{1 \over {5050}}{I_2}\] = I1
\[ \Rightarrow \] \[{{5051} \over {5050}}{I_2}\] = I1
\[ \Rightarrow \] I2 = \[{{5050} \over {5051}}\]I1
Given I2
= \[\alpha \]I1
\[ \therefore \] \[\alpha \] = \[{{5050} \over {5051}}\]