$S=\frac{4^\frac{1}{2012}}{4^\frac{1}{2012}+2}+\frac{4^\frac{2}{2012}}{4^\frac{2}{2012}+2}+...+\frac{4^\frac{2012}{2012}}{4^\frac{2012}{2012}+2}$
Edited by minhdat881439, 14-10-2012 - 15:40.
Tính tổng sau $S=\frac{4^\frac{1}{2012}}{4^\frac{1}{2012}+2}+...$
Started By minhdat881439, 14-10-2012 - 15:39
#1
Posted 14-10-2012 - 15:39
minhdat881439
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Sĩ quan
Tính tổng sau
$S=\frac{4^\frac{1}{2012}}{4^\frac{1}{2012}+2}+\frac{4^\frac{2}{2012}}{4^\frac{2}{2012}+2}+...+\frac{4^\frac{2012}{2012}}{4^\frac{2012}{2012}+2}$
$S=\frac{4^\frac{1}{2012}}{4^\frac{1}{2012}+2}+\frac{4^\frac{2}{2012}}{4^\frac{2}{2012}+2}+...+\frac{4^\frac{2012}{2012}}{4^\frac{2012}{2012}+2}$
- hxthanh likes this
Đừng ngại học hỏi. Kiến thức là vô bờ, là một kho báu mà ta luôn có thể mang theo dể dàng
Trần Minh Đạt tự hào là thành viên VMF
#2
Posted 14-10-2012 - 17:14
hxthanh
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- Hiệp sỹ
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Tín đồ $\sum$
Nhìn sợ thật đấy!
$S=\frac{4^\frac{1}{2012}}{4^\frac{1}{2012}+2}+\frac{4^\frac{2}{2012}}{4^\frac{2}{2012}+2}+...+\frac{4^\frac{2012}{2012}}{4^\frac{2012}{2012}+2}$
$S=\sum\limits_{k=1}^{2012} \dfrac{4^{\frac{k}{2012}}}{4^{\frac{k}{2012}}+2}$
$\Rightarrow S+\dfrac{1}{3}=\sum\limits_{k=0}^{2012} \dfrac{4^{\frac{k}{2012}}}{4^{\frac{k}{2012}}+2}=\sum\limits_{k=0}^{2012} \dfrac{4^{\frac{2012-k}{2012}}}{4^{\frac{2012-k}{2012}}+2}=\sum\limits_{k=0}^{2012} \dfrac{4}{4+2.4^{\frac{k}{2012}}}$
Suy ra:
$2\left(S+\dfrac{1}{3}\right)=\sum\limits_{k=0}^{2012} \dfrac{4^{\frac{k}{2012}}}{4^{\frac{k}{2012}}+2}+\sum\limits_{k=0}^{2012} \dfrac{4}{4+2.4^{\frac{k}{2012}}}=2013$
Do đó: $S=\dfrac{6037}{6}$
$S=\frac{4^\frac{1}{2012}}{4^\frac{1}{2012}+2}+\frac{4^\frac{2}{2012}}{4^\frac{2}{2012}+2}+...+\frac{4^\frac{2012}{2012}}{4^\frac{2012}{2012}+2}$
$S=\sum\limits_{k=1}^{2012} \dfrac{4^{\frac{k}{2012}}}{4^{\frac{k}{2012}}+2}$
$\Rightarrow S+\dfrac{1}{3}=\sum\limits_{k=0}^{2012} \dfrac{4^{\frac{k}{2012}}}{4^{\frac{k}{2012}}+2}=\sum\limits_{k=0}^{2012} \dfrac{4^{\frac{2012-k}{2012}}}{4^{\frac{2012-k}{2012}}+2}=\sum\limits_{k=0}^{2012} \dfrac{4}{4+2.4^{\frac{k}{2012}}}$
Suy ra:
$2\left(S+\dfrac{1}{3}\right)=\sum\limits_{k=0}^{2012} \dfrac{4^{\frac{k}{2012}}}{4^{\frac{k}{2012}}+2}+\sum\limits_{k=0}^{2012} \dfrac{4}{4+2.4^{\frac{k}{2012}}}=2013$
Do đó: $S=\dfrac{6037}{6}$
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