Tính 1+3/2^3+4/2^4+...+100/2^100

Bài 1:

Đặt A = \(1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+\dfrac{5}{2^5}+............+\dfrac{100}{2^{100}}\)

\(\Rightarrow2A=2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+\dfrac{5}{2^4}+...........+\dfrac{100}{2^{99}}\)

\(\Rightarrow2A-A=1+\dfrac{3}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+.....+\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)

\(\Rightarrow A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+......+\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)

\(\Rightarrow2A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+......+\dfrac{1}{2^{98}}-\dfrac{100}{2^{99}}\)

\(\Rightarrow2A-A=2-\dfrac{100}{2^{99}}-\dfrac{1}{2^{99}}+\dfrac{100}{2^{100}}\)

\(\Rightarrow A=2-\dfrac{101}{2^{99}}+\dfrac{100}{2^{100}}\)

\(\Rightarrow A=\dfrac{2^{101}}{2^{100}}-\dfrac{202}{2^{100}}+\dfrac{100}{2^{100}}\)

\(\Rightarrow A=\dfrac{2^{101}-102}{2^{100}}\)

Bài 2:

|4x+3| - x = 15 (*)

TH1: |4x+3|=4x+3 khi 4x+3 \(\ge\) 0 =>4x \(\ge\) -3 => x \(\ge\) \(-\dfrac{3}{4}\)

=> (*) có dạng:

4x+3-x=15

3x+3=15

3x=12

x=4 > \(-\dfrac{3}{4}\) ( thỏa mãn)

TH2: |4x+3|=-4x-3 khi 4x+3 <0 => 4x<-3 => x<\(-\dfrac{3}{4}\)

=> (*) có dạng:

(-4x-3)-x=15

-5x-3=15

-5x=18

x=\(-\dfrac{18}{5}\)< \(-\dfrac{3}{4}\) ( thỏa mãn )

Vậy \(x=4\) và x = \(-\dfrac{18}{5}\) thỏa mãn đề bài.