How many consecutive odd numbers will be needed to obtain the sum of $6^{3}$ ?

6

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5

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4

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3

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Solution

The correct option is A $6$
We know,
$1 = 1 = 1^{3}$ ,
$3 + 5 = 8 = 2^{3}$ ,
$7 + 9 + 11 = 27 = 3^{3}$ ,
$13 + 15 + 17 + 19 = 64 = 4^{3}$ ,
$21 + 23 + 25 + 27 + 29 = 125 = 5^{3}$ ,
$31 + 33 + 35 + 37 + 39 + 41 = 216 = 6^{3}$ .

$∴$ $6$ consecutive odd numbers are required to obtain the sum as $6^{3}$ .
Hence, option $A$ is correct.

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