Download MATH 2283 Solutions to Sample Midterm

MATH 2283
Solutions to Sample Midterm
February 23, 2016
INSTRUCTOR: Anar Akhmedov
1. Prove that for any natural number n,
√
√
1
1
1
2( n + 1 − 1) < 1 + √ + √ + · · · + √ < 2 n.
n
2
3
√
1
1
1
Solution: Let P (n) be the following statement: 2( n + 1−1) < 1+ √ + √ +· · ·+ √ <
n
2
3
√
2 n.
√
One should think of P (n) as two simultaneous inequalities, namely: 2( n + 1 − 1) <
√
1
1
1
1
1
1
1 + √ + √ + · · · + √ and 1 + √ + √ + · · · + √ < 2 n.
n
n
2
3
2
3
√
√
2
2
First notice that P (1) is true: 2( 2 − 1) = √
<
= 1 < 2 1 = 2.
1+1
2+1
√
1
1
Assume P (k) is true, for some natural number k, i.e. Let 2( k + 1 − 1) < 1 + √ + √ +
2
3
√
1
· · · + √ < 2 k. We will prove that P (k + 1) is true.
k
√
√
√
1
1
1
1
1
1+ √ + √ +· · ·+ √ + √
> 2( k + 1−1)+ √
= 2( k + 2−1)−2( k + 2−
k+1
2
3
k
√ k + 1√
√
√
√
√
1
1
2( k + 2 − k + 1)( k + 2 + k + 1)
√
√
= 2( k + 2 − 1) +
+√
=
k + 1) + √
k+1
k+2+ k+1
k+1
√
√
2
1
2
1
√
√
+√
> 2( k + 2−1)− √
+√
=
2( k + 2−1)− √
k+2+ k+1
k+1
k+1+ k+1
k+1
√
2( k + 2 − 1).
√
√
√
√
1
1
1
1
1
1
< 2 k+ √
= 2 k + 1−2( k + 1− k)+ √
=
1+ √ + √ +· · ·+ √ + √
k+1
k+1
2
3√
k √ k√+ 1
√
√
√
2( k + 1 − k)( k + 1 + k)
1
2
1
√
√ +√
√
2 k + 1+
+√
< 2 k + 1− √
<
k+1
k+1
k+1+ k
k+1+ k
√
√
1
2
√
2 k+1− √
+√
= 2 k + 1.
k+1+ k+1
k+1
2. Use mathematical induction to prove that 72n −48n−1 is divisible by 2304 for every natural
number n.
Solution:
Let P (n) be the following statement : 2304 divides s(n), where s(n) = 72n − 48n − 1.
First, notice that s(1) = 72 − 48 − 1 = 0 and 2304 divides 0. This show that P (1) is true.
1
Now let us assume P (k) is true, for some natural number k, i.e. 2304 divides s(k). We will
show P (k + 1) is true.
We have s(k + 1) = 72(k+1) − 48(k + 1) − 1 = 72k 72 − 48(k + 1) − 1 = (72k − 48k − 1)49 +
(48k + 1)49 − 48(k + 1) − 1 = 49s(k) + (48k + 1)49 − 48(k + 1) − 1 = 49s(k) + (49 −
1)48k + 49 − 48 − 1 = 49s(k) + 2304k ≡ 0 (mod 2304). Notice that 2304 divides s(k) by
the inductive assumption. This shows that P (k + 1) is true, if P (k) is true. By induction
P (n) is true for all natural numbers n.
3. The Fibonacci sequence Fn is defined recursively as follows: F1 = F2 = 1, Fn = Fn−1 +Fn−2 .
Prove that every fifth Fibonacci number is divisible by 5.
Solution: First notice that F5 = F4 +F3 = 2F3 +F2 = 4+1 = 5. Assume F5k is divisible by
5 for some natural number k. We have F5(k+1) = F5k+5 = F5k+4 + F5k+3 = 2F5k+3 + F5k+2 =
2(F5k+2 + F5k+1 ) + F5k+2 = 3F5k+2 + 2F5k+1 = 3(F5k+1 + F5k ) + 2F5k+1 = 5F5k+1 + 3F5k .
Thus, F5(k+1) is divisible by 5. By induction F5n is divisible by n for all natural numbers
n.
4. Show that if X is a finite set with n elements, then the number of distinct subsets of X is
2n .
Solution: I have proved this in class on Monday (see lecture notes and textbook, page
24). We used the binary numbers of length n to count the number of subsets of X. You
can also use the induction method. I will do that in class on Wednesday.
5. Prove that for every positive real number x, there is some positive integer n such that
1
0 < < x.
n
Solution: See Theorem 2.4 (textbook, page 33).
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