Download NUMBERS (MA10001): PROBLEM SHEET 2, SOLUTIONS 1. Prove

NUMBERS (MA10001): PROBLEM SHEET 2, SOLUTIONS
1. Prove by induction (and not by any other methods you may know):
1 + 2 + 22 + 23 + · · · + 2n−1 = 2n − 1 for n ∈ N
For instance, 1 + 2 + 22 + 23 + · · · + 2k = 1 + 2(1 + 2 + 22 + 23 + · · · + 2k−1 ) = 1 + 2(2k − 1) =
2k+1 − 1, but there are other ways of doing it. Don’t let forget to start the induction by
checking it for n = 1.
9n − 1 is a multiple of 8 for n ∈ N
Suppose 9k = 8m + 1. Then 9k+1 − 1 = 9(8m + 1) − 1 = 8(9m + 1).
If a > 1 then (1 + a)n > 1 + na2 for n ≥ 3
We get (1 + a)k+1 = (1 + a)(1 + a)k > (1 + a)(1 + ka2 ) by the induction hypothesis. But
(1 + a)(1 + ka2 ) = 1 + a + ka2 + ka3 and since a > 1 and k ≥ 1 we have ka3 > a2 , so
1 + a + ka2 + ka3 > 1 + (k + 1)a2 .
log n < n for n ∈ N
I think it is easiest to say log(k + 1) = log k( k+1
k ) = log k + log
have log k < k so it is enough to show that log
function) that
k+1
k
< e. But
k+1
k
n
X
r=1
=1+
1
k
k+1
k
k+1
k .
By induction we
< 1, that is (since log is an increasing
≤ 2 < e.
n
1
=
for n ∈ N
r(r + 1)
n+1
It is rather a lot easier to write this down if you go from k − 1 to k rather than from k to
Pk−1 1
Pk
1
k + 1. So suppose we know that r=1 r(r+1)
= k−1
r=1 r(r+1) =
k , for some k ≥ 2. Then
Pk−1 1
1
k−1
1
k
r=1 r(r+1) + k(k+1) = k + k(k+1) which simplifies to k+1 .
1
1
1
1
+
+ ··· +
≥ for n ∈ N
n+1 n+2
2n
2
1
1
1
1
1
1
1
k+2 + k+3 + · · · + 2(k+1) = k+1 + k+2 + · · · + 2k − k+1
1
1
1
1
1
1
1
k+1 + 2k+1 + 2(k+1) = 2 − 2(k+1) + 2k+1 > 2 . This is very
1
2k+1
1
2(k+1)
Just write
+
1
2
easy: it just looks
−
complicated, but it isn’t.
2n > (n + 1)2 for n > 5
+
>
This one is easy (some like this can be complicated): 2k+1 = 2 × 2k > 2(k + 1)2 =
2k 2 + 4k + 2 = (k + 2)2 + k 2 − 2, and that is greater than (k + 2)2 as long s k 2 > 2, for
which k ≥ 5 is plenty. Notice that 26 = 64 > 72 = 49 (but 25 = 32 < 62 = 36).
2. Find the sum of all the natural numbers less than 10000 that leave a remainder of 5
when divided by 17.
This should be easy. It is an AP with first term 5 and common difference 17. 588 × 17 =
9996 so there are 588 terms, so the solution is 588 × 5 + 588 × 587 × 17/2 = 2936766.
2
3. What is the value of 23 ?
512 (not 64)
Show that for any t 6= 1,
n Y
n
1+t
2(r−1)
r=1
1 − t2
.
=
1−t
n+1
n
n
2
2
True for n = 1 when it says (1 + t)(1 − t) =
t2 )(1 − t2 ) so
= (1 +
1 − t n. 1 −Qt
Q
2n+1
2n
(r−1)
(r−1)
n
n+1
n
1−t
2
2
(1 + t2 ) = r=1 1 + t2
= 1−t
r=1 1 + t
1−t
1−t (1 + t ) =
What is the value of the left-hand side when t = 1?
Clearly 2n
4. Prove by induction that the sum of the cubes of the first n natural numbers is equal to
the square of the sum of the first n natural numbers, that is
!2
n
n
X
X
r3 =
r .
r=1
r=1
True for n = 1, when both sides are equal to 1.
n+1
X
3
r =
n
X
r3 + (n + 1)3
r=1
r=1
=
n
X
!2
r
+ (n + 1)3
r=1
1 2
n (n + 1)2 + (n + 1)3
4
1
= (n + 1)2 n2 + 4(n + 1)
4
1
= (n + 1)2 (n + 2)2
4
=
Calculate the sum of the cubes of all integers between 30 and 60 inclusive.
P60
P29
We want r=1 r3 − r=1 r3 = 41 602 × 612 − 14 292 × 302 = 152 × (4 × 612 − 292 ) = 3159675
GKS, 2/11/08