If the sum of the first n positive integers is x, then which of the following is the sum of the first 2n even positive integers? Choose the correct option:
A) 2x + 2n^2 B) 3x + 4n^2 C) 4x + 2n D) 4x + n^2 E) 4x + 2n^2
n(n+1)/2 = x   or n^2 + n = 2x (1)   If the series has first 2n numbers, sum of n odd numbers = n^2.   So sum of n even numbers= Sum of first 2n numbers - n^2   In this case we have 2n even numbers so 2n odd numbers will be there, hence a total of 4n first numbers.   Sum of first 4n numbers = 4n(4n+1)/2 -----------(2)   Sum of first 2n odd numbers is (2n)^2 = 4n^2 ---------------- (3)   So sum of 2n even numbers is (2)-(3)   or 4n^2+2n   or 2n^2 + 2 (n^2+n)   or 2n^2 + 2(2x)   or 2n^2 + 4x   ANSW4x + 2n^2R:4x + 2n^2
<result>E</result>
If the sum of the first n positive integers is x, then which of the following is the sum of the first 2n even positive integers?