Chủ đề này có 1 trả lời

[nolunne]

Rút gọn biểu thức sau:

• S=$\frac{3}{2^{2}}+\frac{5}{2^{4}}+\frac{7}{2^{6}}+..+\frac{2n+1}{2^{2n}}$

[DarkBlood]

Rút gọn biểu thức sau:

$S=\frac{3}{2^{2}}+\frac{5}{2^{4}}+\frac{7}{2^{6}}+..+\frac{2n+1}{2^{2n}}$

Ta có:

$4S=3+\frac{5}{2^2}+\frac{7}{2^4}+ \cdot \cdot \cdot + \frac{2n+1}{2^{2(n-1)}}$

 

$\Rightarrow 3S=3+\frac{5-3}{2^2}+\frac{7-5}{2^4}+\cdot \cdot \cdot + \frac{2n+1-[2(n-1)+1]}{2^{2(n-1)}}-\frac{2n+1}{2^{2n}}$

 

$\Rightarrow 3S=3-\frac{2n+1}{2^{2n}}+\frac{2}{2^2}+\frac{2}{2^4}+\cdot \cdot \cdot + \frac{2}{2^{2(n-1)}}$

 

$\Rightarrow 3S=3-\frac{2n+1}{2^{2n}}+\frac{1}{2}+\frac{1}{2^3}+\cdot \cdot \cdot + \frac{1}{2^{2n-3}}$

 

Đặt $M=\frac{1}{2}+\frac{1}{2^3}+\cdot \cdot \cdot + \frac{1}{2^{2n-3}}$

 

Khi đó: $4M=2+\frac{1}{2}+\cdot \cdot \cdot + \frac{1}{2^{2n-5}}$

 

$\Rightarrow 3M=2-\frac{1}{2^{2n-3}}=\frac{2^{2n-2}-1}{2^{2n-3}}$

 

$\Rightarrow M=\frac{2^{2n-2}-1}{3\ .\ 2^{2n-3}}$

 

Do đó: $3S=3-\frac{2n+1}{2^{2n}}+\frac{2^{2n-2}-1}{3\ .\ 2^{2n-3}}$

 

$\Rightarrow 3S=\frac{2^{2n}.9-6n-3+2^{2n+1}-8}{3\ .\ 2^{2n}}$

 

$\Rightarrow S=\frac{2^{2n}.9+2^{2n}.2-11-6n}{9\ .\ 2^{2n}}$

 

Vậy $\boxed{S=\dfrac{11(2^{2n}-1)-6n}{9\ .\ 2^{2n}}}$

Bài viết đã được chỉnh sửa nội dung bởi Hoang Huy Thong: 19-05-2013 - 12:15