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

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 There are 2025 items in 452, after removing 45 square numbers, there are 1980 numbers left, so the square number is removed

Delete the positive integer column, 1,2,3 We get a new sequence of all the complete square numbers in. What is the 2010 term of this sequence?

Because 44 × 44 = 1936 45 × 45 = 2025 46 × 46 = 2116
Therefore, 45 complete square numbers should be deleted from the sequence to 2010
Item 2010 should be 2010 + 45 = 2055

It is known that the function f (x) = X3 + AX2 + BX has extremum at x = - 23 and x = 1. (I) find the value of a and B; (II) find the monotone interval, maximum and minimum of F (x)

(1) F (x) = X3 + AX2 + BX, f ′ (x) = 3x2 + 2aX + B & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; from F ′ (23) = 129-43a + B = 0, f ′ (1) = 3 + 2A + B = 0 & nbsp; & nbsp; & nbsp; & nbsp;, a = - 12, B = - 2 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & n

When we ask the values of a and B, the point (1,3) is the inflection point of the curve y = a * x ^ 3 + b * x ^ 2?

Let f (x) = y, then f (x) = a * x ^ 3 + b * x ^ 2F '(x) = 3A * x ^ 2 + 2B * XF' '(x) = 6A * x + 2B. Because the function has an inflection point at point (1,3), if x = 1 is f' '(x) = 0, that is, 6a + 2B = 0, that is, if point (1,3) is the inflection point of F (x), then f (1) = 3, that is, a + B = 3