(2^(2)+4^(2)+6^(2)+...+20^(2))=?

Given that 1^(2) + 2^(2) + 3^(2) + ... + 20^(2) = 2870 , the value of (2^(2) + 4^(2) + 6^(2) + ... + 40^(2)) is :

Find the sum 5^(2)+ 6^(2) + 7^(2) + ... + 20^(2) .

Given that (1^(2)+2^(2)+3^(2)+..20^(2))=2870, the value of (2^(2)+4^(2)+6^(2)++40^(2)) is 2870 (b) 5740(c)11480(d)28700

5^(2)+6^(2)+7^(2)+.........+20^(2)

tan^(6)20^(@) - 33tan^(2) 20^(@) + 27 tan^(2) 20^(@) + 4=

If (1 + x - 3x^(2))^(10) = a_(0) + a_(1)x + a_(2)x^(2) + ....... + a_(20)x^(20) , then a_(2) + a_(4) + a_(6) + ……. + a_(20) =

tan^(6)20^(@) - 33tan^(4) 20^(@) + 27 tan^(2) 20^(@) + 4=

Let S={1,2,...,20}. A subset B of S is said to be 'nice', if the sum of the elements of B is 203. Then the probability that a randomly chosen subset of S is 'nice' is: (a) 7/(2^20) (b) 5/(2^20) (c) 4/(2^20) (d) 6/(2^20)