Question

How many consecutive odd numbers are needed to make sum as ${13}^3$ ?

• A. $11$
• B. $13$
• C. $15$
• D. $17$

Answer 1

The correct option is B $13$

We know,

$1=1=1^3$,
$3+5=8=2^3$,
$7+9+11=27=3^3$,

$13+15+17+19=64=4^3$,

$21+23+25+27+29=125=5^3$,

$31+33+35+37+39+41=216=6^3$,

$43+45+47+49+51+53+55=343=7^3$,

$57+59+61+63+65+67+69+71=512=8^3$,

$73+75+77+79+81+83+85+87+89=729=9^3$,

$91+93+95+97+99+101+103+105+107+109=1000={10}^3$,

$111+113+115+117+119+121+123+125+127+129+131=1331={11}^3$,

$133+135+137+139+141+143+145+147+149+151+153+155=1728={12}^3$,

$157+159+161+163+165+167+169+171+173+175+177+179+181=2197={13}^3$.

$\therefore$ $13$ consecutive odd numbers are required to obtain the sum as ${13}^3$.
Hence, option $B$ is correct.