Question

Tìm x biết x + \(\frac{3}{2}=1+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2006}}\)

Answer 1

x + $\frac{3}{2}=1+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2006}}$

=>3.(x+\(\frac{3}{2}\))=3.(1+\(\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{2006}}\))

=>3(x+\(\frac{3}{2}\))=3+\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{2005}}\)

=>3(x+\(\frac{3}{2}\))-(x+\(\frac{3}{2}\))=(3+\(1+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\))-($1+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2006}}$

=>2(x+\(\frac{3}{2}\))=3-\(\frac{1}{3^{2006}}\)

=>2(x+\(\frac{3}{2}\))=\(\frac{3^{2007}}{3^{2006}}\)-\(\frac{1}{3^{2006}}\)

=>2(x+\(\frac{3}{2}\))=\(\frac{3^{2007}-1}{3^{2006}}\)

=>x+\(\frac{3}{2}\)=\(\frac{3^{2007}-1}{3^{2006}}:2\)

=>x+\(\frac{3}{2}=\frac{3^{2007}-1}{3^{2006}.2}\)

=>x=\(\frac{3^{2007}-1}{3^{2006}.2}-\frac{3}{2}\)

Answer 2

https://i.imgur.com/JFUf8C9.jpg

Answer 3