College Algebra, 2013a

9.3 Mathematical Induction 679
9.3.2 Selected Answers
n∑
1. Let P (n) be the sentence j 2 =
j=1
1∑
j=1
n(n + 1)(2n +1)
. For the base case, n =1,weget
6
j 2 ? =
(1)(1 + 1)(2(1) + 1)
6
1 2 = 1̌
We now assume P (k) is true and use it to show P (k +1)istrue. Wehave
k∑
j=1
k(k + 1)(2k +1)
6
} {{ }
Using P (k)
∑k+1
j 2 ? (k + 1)((k + 1) + 1)(2(k +1)+1)
=
6
j=1
j 2 +(k +1) 2 ? =
(k +1)(k + 2)(2k +3)
6
+(k +1) 2 ? =
(k +1)(k + 2)(2k +3)
6
k(k + 1)(2k +1) 6(k +1)2
+
6
6
k(k + 1)(2k +1)+6(k +1) 2
6
(k +1)(k(2k +1)+6(k +1))
6
(k +1) ( 2k 2 +7k +6 )
6
(k +1)(k + 2)(2k +3)
6
?
=
?
=
?
=
?
=
=
(k +1)(k + 2)(2k +3)
6
(k +1)(k + 2)(2k +3)
6
(k +1)(k + 2)(2k +3)
6
(k +1)(k + 2)(2k +3)
6
(k +1)(k + 2)(2k +3)
̌
6
By induction,
n∑
j 2 =
j=1
n(n + 1)(2n +1)
6
is true for all natural numbers n ≥ 1.
4. Let P (n) be the sentence 3 n >n 3 . Our base case is n = 4 and we check 3 4 =81and
4 3 =64sothat3 4 > 4 3 as required. We now assume P (k) is true, that is 3 k >k 3 ,and
try to show P (k + 1) is true. We note that 3 k+1 =3· 3 k > 3k 3 and so we are done if
we can show 3k 3 > (k +1) 3 for k ≥ 4. We can solve the inequality 3x 3 > (x +1) 3 using
the techniques of Section 5.3, and doing so gives us x> 3√ 1 ≈ 2.26. Hence, for k ≥ 4,
3−1
3 k+1 =3· 3 k > 3k 3 > (k +1) 3 so that 3 k+1 > (k +1) 3 . By induction, 3 n >n 3 is true for all
natural numbers n ≥ 4.