PROVE BY MATHEMATICAL INDUCTION: (A) 1+2+3+...+N=1/2(N)(N+1) This equality states that the sum of the first n positive integers is equal to 1/2(n)(n+1). (B) 2+4+6+...+2N=N(N+1) (C) 4+8+12+...+4N=2N(N+1) (D) 2 raise to 0 + 2 raise to 1 + 2 raise to 2 +...+ 2 raise to n=2 raise to (n+1) - 1
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