Indução

Mostre, por indução, para todo [latex]n \in \mathbb{N}^*[/latex], que: 

Para n=1:

[latex](1+1) = 2^1\cdot(2\cdot 1-1)[/latex]


Válido.


Assumimos que vale para n = k:


[latex]\left(k+1\right)\cdot\left(k+2 \right )\cdots\left[k+ (k-1)\right ]\cdot\left(k+k \right ) = 2^k\cdot1\cdot3\cdot5\cdots\left(2k-1 \right )[/latex]


Agora queremos demonstrar que a proposição é válida para n = k+1, ou seja:


[latex]\left[\left(k+1\right) +1\right]\cdot\left[\left(k+1\right) +2\right]\cdots\left[\left( k+1\right )+k \right ]\cdot\left[\left(k+1 \right ) + \left(k+1 \right )\right] = 2^{k+1}\cdot1\cdot3\cdot5\cdots\left(2\cdot(k+1)-1 \right )[/latex]


Voltando à nossa expressão:


[latex]\left(k+1\right)\cdot\left(k+2 \right )\cdots\left[k+ (k-1)\right ]\cdot\left(k+k \right ) = 2^k\cdot1\cdot3\cdot5\cdots\left(2k-1 \right )[/latex]



Multiplicando ambos os lados por [latex]\left[\left( k+1\right )+k \right ]\cdot\left[\left(k+1 \right ) + \left(k+1 \right )\right][/latex]:


[latex]\left(k+1\right)\cdot\left(k+2 \right )\cdots\left[k+ (k-1)\right ]\cdot\left(k+k \right )\cdot\left[\left( k+1\right )+k \right ]\cdot\left[\left(k+1 \right ) + \left(k+1 \right )\right] = 2^k\cdot1\cdot3\cdot5\cdots\left(2k-1 \right )\cdot \left[\left( k+1\right )+k \right ]\cdot\left[\left(k+1 \right ) + \left(k+1 \right )\right][/latex]

Passando [latex]\left(k+1\right) [/latex] para o outro lado:


[latex]\left(k+2 \right )\cdots\left[k+ (k-1)\right ]\cdot\left(k+k \right )\cdot\left[\left( k+1\right )+k \right ]\cdot\left[\left(k+1 \right ) + \left(k+1 \right )\right] = 2^k\cdot1\cdot3\cdot5\cdots\left(2k-1 \right )\cdot \left[\left( k+1\right )+k \right ]\cdot\dfrac{\left[\left(k+1 \right ) + \left(k+1 \right )\right]}{\left(k+1\right)}[/latex]

[latex]\left[\left( k+1\right )+1 \right]\cdots\left[\left(k+1 \right )+ (k-2)\right ]\cdot\left[\left( k+1\right )+\left(k-1 \right )\right ]\cdot\left[\left( k+1\right )+k \right ]\cdot\left[\left(k+1 \right ) + \left(k+1 \right )\right] = 2^k\cdot1\cdot3\cdot5\cdots\left(2k-1 \right )\cdot \left[\left( k+1\right )+k \right ]\cdot2[/latex]


[latex]\left[\left( k+1\right )+1 \right]\cdots\left[\left(k+1 \right )+ (k-2)\right ]\cdot\left[\left( k+1\right )+\left(k-1 \right )\right ]\cdot\left[\left( k+1\right )+k \right ]\cdot\left[\left(k+1 \right ) + \left(k+1 \right )\right] = 2^{k+1}\cdot1\cdot3\cdot5\cdots\left(2k-1 \right )\cdot \left[2k+1+(1-1) \right ][/latex]



[latex]\left[\left( k+1\right )+1 \right]\cdots\left[\left(k+1 \right )+ (k-2)\right ]\cdot\left[\left( k+1\right )+\left(k-1 \right )\right ]\cdot\left[\left( k+1\right )+k \right ]\cdot\left[\left(k+1 \right ) + \left(k+1 \right )\right] = 2^{k+1}\cdot1\cdot3\cdot5\cdots\left(2k-1 \right )\cdot \left[2\cdot(k+1)-1 \right ][/latex]



É válida para k+1.


Espero que seja isso 

Valeu! Muito obrigado!