Question

If $1^3+2^3+3^3+.....+9^3=2025$, find the value close to $(0.11{)}^3+(0.22{)}^3+.....+(0.99{)}^3$

• A. $0.2695$
• B. $2.695$
• C. $3.695$
• D. $2.025$

Answer 1

The correct option is B $2.695$

Given $1^3+2^3+3^3+...+9^3=2025$

$S_n={\left(0.11\right)}^3+{\left(0.22\right)}^3+....+{\left(0.99\right)}^3{S}_n={\left(\frac{11}{100}\right)}^3+{\left(\frac{22}{100}\right)}^3+....+{\left(\frac{99}{100}\right)}^3{S}_n={\left(\frac{11}{100}\right)}^3(1^3+2^3+3^3+...+9^3)S_n={\left(\frac{11}{100}\right)}^3\left(2025\right)=2695275\times {10}^{-6}{S}_n=2.695275$