IB HL Math Homework #2: Logs, Binomial Theorem and Induction Solutions
Assigned: 8/29/07, Wednesday
Due: 9/6/07, Thursday
1) (PM pg. 40 # 6(i)) Solve for real x in the equation 4(32x+1) + 17(3x) – 7 = 0.
Solution
4(32x+1) + 17(3x) – 7 = 0
4(3)(32x) + 17(3x) – 7 = 0
12(32x) + 17(3x) – 7 = 0
((4)3x + 7)((3)3x – 1) = 0
((4)3x + 7)(3x+1 – 1) = 0
 3x+1 = 1, so x+1 = 0 and x = -1.
2) ('90 AHSME #23) If x, y > 0, logyx + logxy =
10
x y
, and xy = 144, what is
?
3
2
Solution
Let M = logyx. It then follows that logxy =
ln y
1
1
1



. Substituting,
ln x ln x log y x M
ln y
into the first equation, we have:
1 10

M
3
2
3M  10 M  1  0
(3M  1)( M  3)  0 , so M = 3 or 1/3. Thus, either x = y3 or y = x3.
M
Without loss of generality, assume the latter and substitute into the second equation:
x(x3) = 144, so x4 = 144, and x = 12 , since x > 0. Thus, y = 12 12 , and the answer
x y
12  12 12 13 12 26 3



 13 3 . (The other
to the given question is
2
2
2
2
set of values of x and y lead to this same exact answer.)
3) ('03 AMC-12 B #17) If log(xy3) = 1 and log(x2y) = 1, what is log(xy)?
Solution
Let a = log x and b = log y.
log(xy3) = 1
log(x2y) = 1
log x + log y3 = 1
log x2 + log y = 1
log x + 3log y = 1
2log x + log y = 1
a + 3b = 1
2a + b = 1
-6a – 3b = -3
---------------5a
= -2, so a = .4 and b = .2
Finally, log(xy) = log x + log y = .4 + .2 = .6.
10
2

4) Find the coefficient of x in the expansion of  x 2   .
x

Solution
10
10 10
  2 k 2 10k 10 10 
 2 2
   (2)10k x 3k 10
 x       x ( )
x
x

k 0  k 
k 0  k 
11
To get the coefficient of x11, we need to find the value of k such that
3k – 10 = 11
3k = 21
k=7
Substituting k = 7 for the general term above, we find the coefficient to x11 in the
10 
expansion above is  (2) 3  120(8)  960
7
20
3

5) Find the coefficient of x in the expansion of  x   .
x

Solution
20
20 20
  k 3 20k 20  20 
3

   (3) 20k x 2 k 20
 x       x ( )
x
x

k 0  k 
k 0  k 
To get the coefficient of x, we need to find the (integer) value of k such that
2k – 20 = 1
2k = 21
But by this step we can clearly see that no integral k satisfies this equation. This means
that NONE of the terms in the expansion are an x term. Thus, the desired coefficient is 0.
6) Use induction on n to prove that the following summation is true for all non-negative
integers n:
n
 (i  2)2
i
 (n  1)2 n 1 , for all n  0.
i 0
Solution
(Basis Step) Consider n = 0. In this case,
0
The LHS   (i  2)2 i (0  2)2 0  2; and
i 0
the RHS  (0  1)2 01  2.
Thus, LHS = RHS, so the Basis Step is proved.
(Induction Hypothesis) Consider n = k. We assume the following is true:
k
i
k 1
 (i  2)2  (k  1)2 , for some k  0.
i 0
(Induction Step) Consider n = k + 1. We need to prove
k 1
i
k 11
 (k  2)2 k 2 - - - (1).
 (i  2)2  (k  1  1)2
i 0
k 1
k
i 0
i 0
 (i  2)2 i   (i  2)2 i  (k  1  2)2 k 1 , by the definition of summation
 (k  1)2 k 1  (k  3)2 k 1 , by the Induction Hypothesis
 (k  1  k  3)2 k 1 , factoring
 (2k  4)2 k 1
 2(k  2)2 k 1
 (k  2)2 k  2 , because 2(2 k 1 )  2 k  2
 RHS of (1).
Thus, we proved the Induction Step.
By induction, we proved that the summation identity is true for all n  0.
7) Harmonic numbers Hk , k =1, 2, 3, … are defined by
k
1
Hk   .
i 1 i
Solution
n
Use mathematical induction on n to prove that
H
i 1
i
 (n  1) H n  n
for all n  1.
Use induction on n to prove the assertion.
Base case: n=1: LHS = H1 = 1
RHS = (1+1)H1 - 1 = 2(1) - 1 = 1
Thus, the formula is true for n=1.
Inductive Hypothesis: Assume for an arbitrary value of n=k that
k
H
i 1
i
 (k  1) H k  k
Inductive Step: Prove for n=k+1 that
k 1
H
i 1
k 1
k
H  H
i 1
i
i 1
i
i
 (k  2) H k 1  (k  1)
 H k 1
 (k  1) H k  k  H k 1 , using the inductive hypothesis.
1
 (k  1)( H k 1 
)  k  H k 1 , using the definition of harmonic #s.
k 1
1
 (k  1) H k 1  (k  1) 
 k  H k 1
k 1
 (k  2) H k 1  1  k , combining Hk+1 terms
 (k  2) H k 1  (k  1)
This proves the inductive step using the inductive hypothesis. Thus, we can conclude that
for all integers n  1,
n
H
i 1
i
 (n  1) H n  n .
8) Use induction on n to prove that the following assertion is true for all non-negative
integers n:
 2  1
n 1  n 

  

 n  1
1 0 
 n
n
Solution
 2  1  2  1
  
 , RHS =
Base case n=1: LHS = 
1 0  1 0 
1
1  1  1   2  1

  
 .
 1  1  1  1 0 
Thus, both sides are equal.
Inductive hypothesis: Assume for an arbitrary positive integer n=k that
 2  1
k 1  k 

  

 k  1
1 0 
 k
k
Inductive step: Prove for n=k+1 that
 2  1


1 0 
 2  1


1 0 
k 1
k 1
 (k  1)   k  2  k  1
k  2
  

 
 k 
 k  1  (k  1)  1  k  1
 2  1  2  1
 

 
1 0  1 0 
k
 k  1  k  2  1

 , using the inductive hypothesis.
 
 k  1 1 0 
 k
 2(k  1)  k  (k  1)  k (0) 

= 
 2k  (k  1)  k  0(k  1) 
 k  2  k  1
 , as desired.
= 
 k 
 k 1
 2  1
n 1  n 
  
.
Thus, for all positive integers n, 
 n  1
1 0 
 n
n
9) Using induction on n, prove for all non-negative integers n, 9 | (7n+2 + 52n+1).
Solution
Base case: n=0. 70+2 + 52(0)+1 = 49 + 5 = 54, since 54 = 9(6), we have 9 | 54 and the base
case is true.
Inductive hypothesis: Assume for an arbitrary value of n=k that 9 | (7k+2 + 52k+1).
Equivalently, assume there exists an integer c such that (7k+2 + 52k+1) = 9c.
Inductive step: Prove for n=k+1 that 9 | (7(k+1)+2 + 52(k+1)+1). Equivalently, we must prove
that there exists an integer c' such that (7(k+1)+2 + 52(k+1)+1) = 9c'.
(7(k+1)+2 + 52(k+1)+1) = 7(7k+2) + 52k+3
= 7(7k+2) + (25)52k+1
= 7(7k+2) + (7+18)52k+1
= 7(7k+2) + (7)52k+1+(18)52k+1
= 7(7k+2 + 52k+1)+(2)(9)52k+1
= 7(9c)+(2)(9)52k+1, using the inductive hypothesis.
= 9(7c+(2)52k+1), proving the inductive step, since c and k are
integers.