b) P= 3 7.5 + 3 7.5^ 2 + 3 7.5^ 3 +...+ 3 7.5^ 2006 giúp mình với ạ
2 câu trả lời
` Huy `
` P=37 . 5 + 37.5^2 + 37.5^3 + ... + 37.5^{2006} `
` => P = 37 . (5+5^2+5^3+...+5^{2006}) `
` M = 5+5^2+5^3+...+5^{2006} `
` =>5M = 5^2+5^3+5^4+...+5^{2007} `
` => 5M - M = (5^2+5^3+5^4+...+5^{2007}) - (5 + 37.5^2 + 37.5^3 + ... + 37.5^{2006} ) `
` => 4M = 5^{2007} - 5 `
` => M= \frac{5^{2007}-5}{4} `
` => P=37 . \frac{5^{2007}-5}{4} = \frac{37.(5^{2007}-5)}{4} `
Đáp án:
$P=\dfrac{21.5^{2007}-105}4$.
Giải thích các bước giải:
$\begin{array}{l}P=3.7.5+3.7.5^2+3.7.5^3\ +\,.\!.\!.+\ 3.7.5^{2006}\\\Rightarrow P=3.7.(5+5^2+5^3\ +\,.\!.\!.+\ 5^{2006})\\\Rightarrow\dfrac P{3.7}=5+5^2+5^3\ +\,.\!.\!.+\ 5^{2006}\\\Rightarrow\dfrac{5P}{21}=5.(5+5^2+5^3\ +\,.\!.\!.+\ 5^{2006})\\\Rightarrow \dfrac{5P}{21}=5^2+5^3+5^4\ +\,.\!.\!.+\ 5^{2007}\\\Rightarrow\dfrac{5P}{21}-\dfrac P{21}=(5^2+5^3+5^4\ +\,.\!.\!.+\ 5^{2007})-(5+5^2+5^3\ +\,.\!.\!.+\ 5^{2006})\end{array}\\\Rightarrow\dfrac{4P}{21}=5^{2007}-5\\\Rightarrow \dfrac P{21}=\dfrac{5^{2007}-5}4\\\Rightarrow P=\dfrac{21.(5^{2007}-5)}4\\\Rightarrow P=\dfrac{21.5^{2007}-105}4$
Vậy $P=\dfrac{21.5^{2007}-105}4$.
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