Question

tính tổng S = 1+ 5/7+5/7^2+5/7^3+....+5/7^55

Answer 1

Ta có: \( 7S=7+5+\frac{5}{7}+\frac{5}{ 7^{2}}+...+\frac{5}{7^{54}}\) 

\(\Rightarrow 7S-S= (7+5+\frac{5}{7}+\frac{5}{ 7^{2}}+...+\frac{5}{7^{54}})-( 1+\frac{5}{7}+\frac{5}{ 7^{2}}+...+\frac{5}{7^{55}})\)

\(\Rightarrow 6S=(7+5-1)+(\frac{5}{7}-\frac{5}{7})+...+(\frac{5}{7^{54}}-\frac{5}{7^{54}})-\frac{5}{7^{55}}\)

\(\Rightarrow 6S=11-\frac{5}{7^{55}}\)

Vậy \(S=\frac{11.7^{55} -30}{6.7^{55}}\)

 

Answer 2

• S= 1+ 5/7+5/7^2+5/7^3+....+5/7^55

• 7S= 7+ 5+ 5/7+5/7^2+5/7^3+....+5/7^54
• 7S-S= (7+ 5+ 5/7+5/7^2+5/7^3+....+5/7^54)-(1+ 5/7+5/7^2+5/7^3+....+5/7^55)
• 6S= 7+5- 1- 7^55
• 6S= 11- 7^55
• S= 11- 7^55 / 6