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7
1.13
(i)
Prove that 2
n
>
n
3
for all
n
≥
10.
Solution.
Base step
.2
10
=
1024
>
10
3
=
1000. (Note that
2
9
=
512
<
9
3
=
729.)
Inductive step
Note that
n
≥
10 implies
n
≥
4. The inductive
hypothesis is 2
n
>
n
3
; multiplying both sides by 2 gives
2
n
+
1
=
2
−
2
n
>
2
n
3
=
n
3
+
n
3
≥
n
3
+
4
n
2
=
n
3
+
3
n
2
+
n
2
>
n
3
+
3
n
2
+
4
n
=
n
3
+
3
n
2
+
3
n
+
n
≥
n
3
+
3
n
2
+
3
n
+
1
=
(
n
+
1
)
3
.
(ii)
Prove that 2
n
>
n
4
for all
n
≥
17.
Solution.
Base step
.2
17
=
131
,
072
>
17
4
=
83
,
521. (Note that
16
4
=
(
2
4
)
4
=
2
16
.)
Inductive step
. Note that
n
≥
17 implies
n
≥
7. The inductive
hypothesis is 2
n
>
n
4
; multiplying both sides by 2 gives
2
n
+
1
=
2
−
2
n
>
2
n
4
=
n
4
+
n
4
≥
n
4
+
5
n
3
≥
n
4
+
4
n
3
+
n
3
≥
n
4
+
4
n
3
+
7
n
2
≥
n
4
+
4
n
3
+
6
n
2
+
n
2
≥
n
4
+
4
n
3
+
6
n
2
+
5
n
≥
n
4
+
4
n
3
+
6
n
2
+
4
n
+
1
=
(
n
+
1
)
4
.
1.14
Around 1350, N. Oresme was able to sum the series
∑
∞
n
=
1
n
/
2
n
by dis
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Unformatted text preview: secting the region in Figure 1.4 in two ways. Let A n be the vertical rectangle with base 1 2 n and height n , so that area ( A n ) = n / 2 n , and let B n be horizontal rectangle with base 1 2 n + 1 2 n + 1 + ··· and height 1. Prove that ∑ ∞ n = 1 n / 2 n = 2. Solution. You may assume that ∑ ∞ n = ar n = a /( 1 − r ) if 0 ≤ r < 1. Now compute the area using ∑ A n = ∑ B n ....
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.
 Fall '11
 KeithCornell

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