Explanation
We know, a perfect cube has multiples of 3 as powers of prime factors.
Here, number of 3's is 3 and number of 37's is 1.
Given number 216 is the cube of 6, i.e. 63=216.
We know,
1=1=13,
3+5=8=23,
7+9+11=27=33,
13+15+17+19=64=43,
21+23+25+27+29=125=53,
31+33+35+37+39+41=216=63.
∴ The sum of six consecutive odd numbers are to form 216 is 31+33+35+37+39+41=216.
Hence, option D is correct.
Given number 64 is the cube of 4, i.e. 43=64.
13+15+17+19=64=43.
∴ The sum of four consecutive odd numbers are to form 64 is 13+15+17+19=64.
Hence, option B is correct.
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