Compute the values of the product 1 + 1 1 1 + 1 2 1 + 1 3 1 + 1 n for small values of n in order to conjecture a general
Posted: Mon Jul 11, 2022 12:03 pm
Compute the values of the product 1 + 1 1 1 + 1 2 1 + 1 3 1 + 1n for small values of n in order to conjecture a general formulafor the product. Fill in the blank with your conjecture. 1 + 1 1 1+ 1 2 1 + 1 3 1 + 1 n = Correct: Your answer is correct. Prove yourconjecture by mathematical induction. Proof (by mathematicalinduction): Let P(n) be the equation 1 + 1 1 1 + 1 2 1 + 1 3 1 + 1n = Incorrect: Your answer is incorrect. . We will show that P(n)is true for every integer n ≥ 1. Show that P(1) is true: SelectP(1) from the choices below. 1 + 1 1 = 1 + 1 P(1) = 1 P(1) = 1 + 11 1 + 1 1 1 + 1 2 1 + 1 3 1 + 1 1 = 1 + 2 + 3 Correct: Your answeris correct. The selected statement is true because both sides ofthe equation equal the same quantity. Show that for each integer k≥ 1, if P(k) is true, then P(k + 1) is true: Let k be any integerwith k ≥ 1, and suppose that P(k) is true. The left-hand side ofP(k) is 1 + 1 1 1 + 1 2 1 + 1 3 Incorrect: Your answer isincorrect. and the right-hand side of P(k) is Correct: Your answeris correct. . [The inductive hypothesis is that the two sides ofP(k) are equal.] We must show that P(k + 1) is true. The left-handside of P(k + 1) is 1 + 1 1 1 + 1 2 1 + 1 3 Incorrect: Your answeris incorrect. and the right-hand side of P(k + 1) is Incorrect:Your answer is incorrect. . After substitution from the inductivehypothesis, the left-hand side of P(k + 1) becomes ---Select---Incorrect: Your answer is incorrect. · 1 + 1 k + 1 . When theleft-hand and right-hand sides of P(k + 1) are simplified, theyboth can be shown to equal Correct: Your answer is correct. . HenceP(k + 1) is true, which completes the inductive step.
Compute the values of the product (¹ + ²) (¹ + 2) (¹ + 3) --- (¹ + 2) for small values of n in order to conjecture a general formula for the product. Fill in the blank with your conjecture (¹ + ²) (¹ + ¹) (¹ + ³ ) · · · (¹ + ² ) - | *+1 Prove your conjecture by mathematical induction. Proof (by mathematical induction): Let P(n) be the equation (1+ Show that P(1) is true: Select P(1) from the choices below. @ 1+1=1+1 OP(1) - 1 OM(1)-141 |0 (¹ + ¹ ) ( ¹ + ¹) (¹ + ¹ ) ( ¹ + ÷ ) = ¹+2+ 3 on ( ¹ + ¹ ) (¹ + ²¹) ( ¹ + ¹ ) · · · ( ¹ + ¹ ) -¯¯ ✓ x [The inductive hypothesis is that the two sides of P(X) are equal.] We must show that P(x + 1) is true. The left-hand side of P + 1) is (1 + 1)( ¹ + ²) (¹ + ) · ( --Select--x ).(₁¹. We will show that P(n) is true for every integer n 2 1. The selected statement is true because both sides of the equation equal the same quantity. Show that for each integer k 2 1, if P(k) is true, then P(x + 1) is true: Let & be any integer with & 2 1, and suppose that P(X) is true. The left-hand side of P(x) is (1 + 1)(1+²) (¹ + 3 ) · · ( [ k+1 and the right-hand side of P(k+ 1) is 1). When the left-hand and right-hand sides of P(x + 1) are simplified, they both can be shown to equal +2 X and the right-hand side of P(k) is . After substitution from the inductive hypothesis, the left-hand side of P(k+ 1) becomes Hence P(+1) is true, which completes the inductive step.
Compute the values of the product (¹ + ²) (¹ + 2) (¹ + 3) --- (¹ + 2) for small values of n in order to conjecture a general formula for the product. Fill in the blank with your conjecture (¹ + ²) (¹ + ¹) (¹ + ³ ) · · · (¹ + ² ) - | *+1 Prove your conjecture by mathematical induction. Proof (by mathematical induction): Let P(n) be the equation (1+ Show that P(1) is true: Select P(1) from the choices below. @ 1+1=1+1 OP(1) - 1 OM(1)-141 |0 (¹ + ¹ ) ( ¹ + ¹) (¹ + ¹ ) ( ¹ + ÷ ) = ¹+2+ 3 on ( ¹ + ¹ ) (¹ + ²¹) ( ¹ + ¹ ) · · · ( ¹ + ¹ ) -¯¯ ✓ x [The inductive hypothesis is that the two sides of P(X) are equal.] We must show that P(x + 1) is true. The left-hand side of P + 1) is (1 + 1)( ¹ + ²) (¹ + ) · ( --Select--x ).(₁¹. We will show that P(n) is true for every integer n 2 1. The selected statement is true because both sides of the equation equal the same quantity. Show that for each integer k 2 1, if P(k) is true, then P(x + 1) is true: Let & be any integer with & 2 1, and suppose that P(X) is true. The left-hand side of P(x) is (1 + 1)(1+²) (¹ + 3 ) · · ( [ k+1 and the right-hand side of P(k+ 1) is 1). When the left-hand and right-hand sides of P(x + 1) are simplified, they both can be shown to equal +2 X and the right-hand side of P(k) is . After substitution from the inductive hypothesis, the left-hand side of P(k+ 1) becomes Hence P(+1) is true, which completes the inductive step.