Let n be a complete square number and n be a cubic number, then the minimum positive value of n is zero

Let n be a complete square number and n be a cubic number, then the minimum positive value of n is zero

N △ 2 is a complete square number, so 2 is the prime factor of N, n △ 3 is a cubic number, and at least three of 2 n △ 3 are cubic numbers, so 3 is the prime factor of N. n △ 2 is a complete square number, and at least four of 3 N in n have the minimum value of 3 ^ 4 + 2 ^ 3 = 648n, which can be divided by 2 and 3, so n = 2 ^ m * 3 ^ n is the minimum, and there is no other factor, because n / 2 =

Cube of 1 + cube of 2 + ······ + cube of n = (1 + 2 + 3 ··· + n) squared

By mathematical induction, it is proved that when n = 1 proposition holds, suppose n = k proposition 1 ^ 3 + 2 ^ 3 + 3 ^ 3 +. + K ^ 3 = (K (K + 1) / 2) ^ 2 holds. When n = K + 1, 1 ^ 3 + 2 ^ 3 + 3 ^ 3 +. + K ^ 3 + (K + 1) ^ 3 = (K (K + 1) / 2) ^ 2 + (K + 1) ^ 3 = ((K + 1) * (K + 1 + 1) / 2) ^ 2. In conclusion, proposition 1 ^ 3 + 2 ^ 3 + 3 +. + K ^ 3 = (n (n + 1

Delete positive integer sequence 1, 2, 3 All the complete square numbers in, get a new sequence, the new sequence of 2003 is ()
A. 2048B. 2049C. 2050D. 2051

These numbers can be written as: 12, 2, 3, 22, 5, 6, 7, 8, 32 There are 2K positive integers between the k th square number and the K + 1 th square number, and the sequence 12, 2, 3, 22, 5, 6, 7, 8, 32 452 has a total of 2025 items. After removing 45 square numbers, there are 1980 numbers left. Therefore, after removing the square number, the 2003 item should be the 23rd number after 2025, which is the 2048 item of the original sequence, which is 2048. Therefore, select: a