Toán nâng cao 6!

P

phamhuy20011801

Đặt A= (đề)
\Rightarrow $4A=2^3+ 2^5+2^7+2^9+...+2^{2013}$
$3A=4A-A=2^3+2^5+2^7+...+2^{2013}-(2+2^3+2^5+...+2^{2011})$
$3A=2^{2013}-2$
$A=\dfrac{2^{2013}-2}{3}$
 
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C

chaugiang81

Đặt A= (đề)
\Rightarrow $4A= 2^5+2^7+2^9+...+2^2013$
$3A=4A-A=2^5+2^7+...+2^2013-(2^3+2^5+...+2^2011)$
A=$\frac{2^2013-8}{3}$
bạn ơi, mk nghĩ 4A= ( 2 + 2^3 +2^5 +2^7 +...+ 2^2011) 2^2 \Leftrightarrow 2^3 + 2^5 + 2^7 + 2^9 +...+ 2^2013
\Rightarrow 4A - A= (2^3 + 2^5 + 2^7 + 2^9 +...+ 2^2013) - ( 2 + 2^3 +2^5 +2^7 +...+ 2^2011) \Leftrightarrow 3A = 2 ^ 2013 - 2 \Rightarrow A = (2 ^ 2013 - 2) / 3
đúng khong bạn
 
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