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PROBLEM #4, page 105
1 + 3 + 6 + ... + n(n+1)/2 = n(n+1)(n+2)/6
Use mathematical induction to prove that 1 + 3 + 6 + ... + n(n+1)/2 = n(n+1)(n+2)/6 whenever n is a positive integer.
SOLUTION
Let P(n) denote the proposition "1 + 3 + 6 + ... + n(n+1)/2 = n(n+1)(n+2)/6"
Basis step - Show that P(1) is true
P(1) is true since 1=1*2/2.
Inductive step - Show that if P(n) is true, then P(n+1) is true.
1 + 3 + 6 + ... + n(n+1)/2 = n(n+1)(n+2)/6
1 + 3 + 6 + ... + (n+1)(n+2)/2 = (n+1)(n+2)(n+3)/6
To do this, start with the left side of P(n+1) and modify to equivalent statements until we get the right side of P(n+1).
1 + 3 + 6 + ... + (n+1)(n+2)/2 =
1 + 3 + 6 + ... + n(n+1)/2 + (n+1)(n+2)/2 [just showing back one term in what was ...ed]
n(n+1)(n+2)/6 + (n+1)(n+2)/2 [replacing the left side of P(n) with the right side]
(n+1)(n+2)[n/6+1/2] [factoring out the common terms]
(n+1)(n+2)[n/6+3/6] [creating common denominator]
(n+1)(n+2)[(n+3)/6] [addition]
(n+1)(n+2)(n+3)/6 [association rule]
Thus, the induction step is complete.
Therefore, by the principle of mathematical induction, it has been shown that 1 + 3 + 6 + ... + n(n+1)/2 = n(n+1)(n+2)/6 whenever n is a positive integer
Now, let's see how you do on a couple of problems working with a group. Solve problems 2 and 7 in your groups.
PROBLEM #2, page 105
2 + 4 + 6 + ... + 2n = n(n+1)
Use mathematical induction to prove that 2 + 4 + 6 + ... + 2n = n(n+1) whenever n is a positive integer.
SOLUTION
Let P(n) denote the proposition "2 + 4 + 6 + ... + 2n = n(n+1)"
Basis step - Show that P(1) is true
P(1) is true since 2=1*2.
Inductive step - Show that if P(n) is true, then P(n+1) is true.
2 + 4 + 6 + ... + 2n = n(n+1)
2 + 4 + 6 + ... + 2(n+1) = (n+1)(n+2)
To do this, start with the left side of P(n+1) and modify to equivalent statements until we get the right side of P(n+1).
2 + 4 + 6 + ... + 2(n+1) =
2 + 4 + 6 + ... + 2n + 2(n+1) [just showing back one term in what was ...ed]
n(n+1) + 2(n+1) [replacing the left side of P(n) with the right side]
(n+1)[n+2] [factoring out the common term]
Thus, the induction step is complete.
Therefore, by the principle of mathematical induction, it has been shown that 2 + 4 + 6 + ... + 2n = n(n+1) whenever n is a positive integer
PROBLEM #7, page 105
12 + 22 + ... n2 = n(n+1)(2n+1)/6
Use mathematical induction to prove that 12 + 22 + ... n2 = n(n+1)(2n+1)/6 whenever n is a positive integer.
SOLUTION
Let P(n) denote the proposition "12 + 22 + ... n2 = n(n+1)(2n+1)/6"
Basis step - Show that P(1) is true
P(1) is true since 12=1=1*2*(2*1+1)/6.
Inductive step - Show that if P(n) is true, then P(n+1) is true.
12 + 22 + ... n2 = n(n+1)(2n+1)/6
12 + 22 + ... (n+1)2 = (n+1)(n+2)(2(n+1)+1)/6= (n+1)(n+2)(2n+3)/6
To do this, start with the left side of P(n+1) and modify to equivalent statements until we get the right side of P(n+1).
12 + 22 + ... (n+1)2 =
12 + 22 + ... n2 + (n+1)2 = [just showing back one term in what was ...ed]
n(n+1)(2n+1)/6 + (n+1)2 = [replacing the left side of P(n) with the right side]
(n+1)[n(2n+1)/6+(n+1)] [factoring out the common terms]
(n+1)[n(2n+1)/6+6(n+1)/6] [creating common denominator]
(n+1)[(2n2 + n + 6n + 6)/6] [addition and factoring]
(n+1)[(n+2)(2n+3)/6] [factoring]
(n+1)(n+2)(2n+3)/6 [associative rule]
Thus, the induction step is complete.
Therefore, by the principle of mathematical induction, it has been shown that 12 + 22 + ... n2 = n(n+1)(2n+1)/6 whenever n is a positive integer