Question

Express ${11}^3$ as the sum of consecutive odd numbers.

• A. 66+77+88+99+110+121+132+143+154+165+176
• B. 111+113+115+117+119+121+123+125+127+129+131
• C. 121+123+125+127, 129+131+133+135+137+139+141
• D. 101+103+105+107, 109+111+113+115+117+119+121

Answer 1

The correct option is B 111+113+115+117+119+121+123+125+127+129+131

Here, the sum of 11 consecutive odd numbers should be equal to ${11}^3$.
To find the odd numbers whose sum is ${11}^3$, we first divide ${11}^3$ by 11.
$\frac{{11}^3}{11}={11}^2=121$
Since 121 is an odd number, it would be the middle number of the series.
The other odd numbers would be 5 odd numbers before and after 121.
Thus, ${11}^3$ can be expressed as: 111+113+115+117+119+121+123+125+127+129+131

So, the option (b) is correct.