Question

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

• A. $0.2695$
• B. $0.3695$
• C. $2.695$
• D. $3.695$

Answer 1

Answer: (C)

$2.695$

$(0.11{)}^3+(0.22{)}^3+.........+(0.99{)}^3$

$=(0.11\times 1{)}^3+(0.11\times 2{)}^3+(0.11\times 3{)}^3+............+(0.11\times 9{)}^3$

$=\left(0.11{)}^3[1^3+2^3+3^3+.......+9^3]\right(\text{Taking 0.11 as common})$

$=(0.11{)}^3[1+8+27+64+125+216+343+512+729]$

$=0.001331\times 2025=2.695275=2.695$ (approx)