Prove that the sum of a nite arithmetic series equals the product of the number of terms and half the sum of the two extreme terms; in symbols, this reads (a + d) + (a +2d) + (a +3d) +... [(a+d) + (a +nd)]. 2 +(a+nd) = n Use the result of part (a) to con rm the identities and 1+4+7+ +(3n-2): n(3n-1) 2 1+3+5++ (2n-1)=n.