Any method to find How many (& what, if possible)perfect squares <= N (any given no.)?

Answer 1

Well, the perfect squares are (in order) 0^2, 1^2, 2^2, 3^2, 4^2, ...

If we look at the square root of N, that should give us an idea of how many squares are less than it. For example, if the square root of N is between 7 and 8, then we know that there are eight perfect squares less than or equal to N (they are 0^2, 1^2, 2^2, ..., 7^2).

So to find the number of perfect squares less than or equal to N, we just need to take the next integer greater than the square root of N. Try this with a couple numbers and you'll see why it works.

An easy way to find perfect squares is to just add up consecutive odd integers:
1
1 + 3 = 4
1 + 3 + 5 = 9
1 + 3 + 5 + 7 = 16

These will give you all the perfect squares (except 0, which is special in some sense).

Answer 2

I'm assuming that your definition of 'perfect square' is (something like) 'any number that can be expressed as n^2 where n is a whole number'
ie 4, 9, 16, 25 are all perfect squares.

To find out how many perfect aquares that are less than N, simply take the square root of N. eg root(127) = 11.26;
So, the highest square number that is less than 127 is 11x11=121, and we know there are 11 square numbers up to 127.
These are fairly easy to list:
1^2=1
2^2=4
3^2=9
...
10^2=100
11^2=121

Answer 3

Since the series 1+3+5+7+.....+2n-1.... = n^2
Then if n=3 , you get 1, 4 and 9 as perfect squares =< n.

If n = 20 then n^2=400 and 2n-1 = 39.
Thus the 19th term would be 37 and its value would be
400-39 =361 = 19^2.

Does this help at all?