Recitation 3 2 Problem: A geometric sum Perhaps you encountered this classic formula in school 1+r+r2+r3+.+rn Use induction to prove that it is correct for all real values r+1 Prepare a complete, careful solution. You'll be passing your proof to another group for " construc- tive criticism"! Solution Proof. We use induction. Let P(n) be the proposition that the following equation holds for allr≠1: 1+r+2+r3+..+=1-r Base case: P(O)is true, because both sides of the equation are equal to 1 P(n)is true, where n denotes an arbitrary natural number. We can reason as follows 9F Inductive step: We must show that P(n) implies P(n+1) for all n E N. So assume 1+r+r2+r3 +(1-r) - rn+2 1 The first equation follows from the assumption P(n), and the remaining steps are simpl fications. This proves that P(n+ 1) is also true. Therefore, P(n) implies P(n+ 1)for all mEN. By the principle of induction, P(n)is true for all nE N Note: You may have encountered a different proof of this formula. We'l write down a sequence of equations and then explain the reasoning S=1+r+r2+r3+ rS=r+r2+r3+..+rn+1 s-rS We define S on the first line, multiply by r to get the second equation, subtract the second equation from the first to get the third, and then solve for S. This gives the formula above This argument is great! It is a derivation of the formula rather than just a verification But, at some level, weve only hidden the use of induction, since the operations were doing on n-term sums are justified using you guessed it-inductionRecitation 3 3 2 Problem: A Geometric Sum Perhaps you encountered this classic formula in school: 3 1 + r + r 2 + r + . . . + r n = 1 − rn+1 1 − r Use induction to prove that it is correct for all real values r �= 1. Prepare a complete, careful solution. You’ll be passing your proof to another group for “construc￾tive criticism”!’ Solution. Proof. We use induction. Let P(n) be the proposition that the following equation holds for all r = 1 � : 1 − rn+1 3 1 + r + r 2 + r + . . . + r n = 1 − r Base case: P(0) is true, because both sides of the equation are equal to 1. Inductive step: We must show that P(n) implies P(n + 1) for all n ∈ N. So assume that P(n) is true, where n denotes an arbitrary natural number. We can reason as follows: 3 n+1 1 + r + r 2 + r + . . . + r n + r n+1 = 1 − rn+1 + r 1 − r 1 − rn+1 + (1 − r) · rn+1 = 1 − r 1 − rn+2 = 1 − r The first equation follows from the assumption P(n), and the remaining steps are simpli­ fications. This proves that P(n + 1) is also true. Therefore, P(n) implies P(n + 1) for all n ∈ N. By the principle of induction, P(n) is true for all n ∈ N. Note: You may have encountered a different proof of this formula. We’ll write down a sequence of equations and then explain the reasoning. 3 S = 1 + r + r 2 + r + . . . + r n 3 n+1 rS = r + r 2 + r + . . . + r S − rS = 1 − r n+1 1 − rn+1 S = 1 − r We define S on the first line, multiply by r to get the second equation, subtract the second equation from the first to get the third, and then solve for S. This gives the formula above! This argument is great! It is a derivation of the formula rather than just a verification. But, at some level, we’ve only hidden the use of induction, since the operations we’re doing on n­term sums are justified using— you guessed it— induction