giải giúp mình với ạ P= 3 /7.5 + 3/ 7.5^ 2 + 3 /7.5^ 3 +...+ 3 /7.5^ 2006

Đáp án:

$P=\dfrac{3.5^{2007}-15}{28}$.

Giải thích các bước giải:

$P=\dfrac37\!\cdot\!5+\dfrac37\!\cdot\!5^2+\dfrac37\!\cdot5^3\ +\,.\!.\!.+\ \dfrac37\!\cdot\!5^{2006}\\\Rightarrow P=\dfrac37\bigg(5+5^2+5^3\ +\,.\!.\!.+\ 5^{2006}\bigg)\\\Rightarrow P:\dfrac37=5+5^2+5^3\ +\,.\!.\!.+\ 5^{2006}\\\Rightarrow \dfrac{7P}3=5+5^2+5^3\ +\,.\!.\!.+\ 5^{2006}\\\Rightarrow\dfrac{5.7P}3=5.(5+5^2+5^3\ +\,.\!.\!.+\ 5^{2006})\\\Rightarrow\dfrac{35P}3=5^2+5^3+5^4\ +\,.\!.\!.+\ 5^{2007}\\\Rightarrow\dfrac{35B}3-\dfrac{7P}3=(5^2+5^3+5^4\ +\,.\!.\!.+\ 5^{2007})-(5+5^2+5^3\ +\,.\!.\!.+\ 5^{2006})\\\Rightarrow\dfrac{28P}3=5^{2007}-5\\\Rightarrow\dfrac P3=\dfrac{5^{2007}-5}{28}\\\Rightarrow P=\dfrac{3.(5^{2007}-5)}{28}\\\Rightarrow P=\dfrac{3.5^{2007}-15}{28}$

`P=3/7 . 5 +3/7 . 5^2+...+3/7 . 5^{2006}`

`=3/7 . (5+5^2+...+5^{2006})`

Đặt `A=5+5^2+...+5^{2006}`

`->5A=5^2+5^3+...+5^{2007}`

`->5A-A=5^{2007}-5`

`-> 4A=5^{2007}-5`

`->A=(5^{2007}-5)/4`

`-> P = 3/7 . (5^{2007}-5)/4`

`-> P = (3.5^{2007}-15)/28`

Vậy `P = (3.5^{2007}-15)/28`