What is the 100th term of the sequence 2, 3, 5, 6, 7, 10, 11, ... which consists of all of the positive integers that are neither perfect squares nor perfect cubes?

112

Step-by-step explanation:

The given sequence is

2, 3, 5, 6, 7, 10, 11, ...

It consists of all of the positive integers that are neither perfect squares nor perfect cubes.

We need to find the 100th term of the sequence.

Squares = 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121

Cubes = 1, 8, 27, 64, 125

Squares and cubes from 1 to 100 are 1, 4, 8, 9, 16, 25, 27, 36, 49, 64, 81, 100.

Total 12 numbers are squares or cubes.

So, the 100th term of this sequence is

100 + 12 = 112

Therefore, the 100th term is 112.

112

Step-by-step explanation:

There are 10 squares in the interval [1, 100], and there are 4 cubes in that interval, two of which are also squares (1, and 64). Hence 12 numbers must be added to the list to make up for the ones that are stricken. The 100th number is 112.