" 49."(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 :

Given that (1^(2)+2^(2)+3^(2)+......+10^(2))=385, the value of (2^(2)+4^(2)+6^(2)+....+20^(2)) is equal to

If 1^(2) + 2^(2) + 3^(3)+ 4^(2) + …….. + 10^(2) = 385 then find the value of 2^(2) + 4^(2) + 6^(2) + ………. + 20^(2)

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

a^(2)-49b^(2)

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

Find (49(1)/(2))^(2)

If t_(n) denotes the nth term of the series 2+3+6+11+18+... then t_(50)49^(2)-1b .49^(2)c.50^(2)+1d*49^(2)+2

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