CMSC 203 - Homework Assignment 3 - Due April 24, 2008

Name ___Solution Key_____CMSC 203 - Homework Assignment 3 - Due April 24, 20083. Using Mathematical Induction, prove for all integers n > 1,n + 1∑ i ⋅ 2 i = n ⋅ 2 n + 2 + 2i = 1Proof: (Induction)Basis: Since the statement specifies n > 1, we see that at n = 1, the LHS = 1(2 1 ) + 2(2 2 ) = 2 + 8 = 10, and theRHS = 1(2 1+2 ) + 2 = 2 3 + 2 = 8 + 2 = 10, so LHS = RHS.However, if we consider the smallest allowable value for n here, we see that n = 0 is also a viable basis value.In this case, the LHS = 1(2 1 ) = 2 and the RHS = 0(2 0+2 ) + 2 = 2, so again LHS = RHS, and the Basis is valid.Inductive: Assume the statement is true for n = k, that is,( k + 1) + 1Nowk + 1∑ i ⋅ 2 i = ( k + 1) ⋅ 2 ( k + 1)+ 2 + 2 = ( k + 1) ⋅ 2 k + 3 + 2 .i = 1k + 2∑i = 1i 2 i ⋅k + 1= (2k + 2)2 k+2 + 2 = (k + 1)(2)(2 k+2 ) + 2 = (k + 1)2 k+3 + 2. QED∑ i ⋅ 2 i = k ⋅ 2 k + 2 + 2 . We want to show thati = 1= i ⋅ 2 i + i ⋅ 2 i = k ⋅ 2 k + 2 + 2 + ( k + 2) ⋅ 2 k + 2 = ( k + ( k + 2)) ⋅ 2 k 2 +∑i = 1k + 2∑i = k+2+ 2