5.4 mathematical induction t

To use induction to verify infinitely many statements
S1, S2, S3,.. , always
0. declare that an induction argument is used,
1. verify the first statement S1 is true,
2. assuming the N'th statement SN is true, use this to
verify that as a consequence, SN+1 must also be true.
Example A.
Verify that (2k – 1) = n2 for all natural numbers n.
k=1
n
We will use induction. (declaring the methodology)
I. Verify its true when n = 1. (i.e S1 is true )
If n = 1, we've (2k – 1) = 1 = 12
. Done.k=1
1
Mathematical Induction
These steps are established format when invoking the
mathematical induction and they should be followed.

II. Assume the formula works for n = N.
(This means that (2k – 1) = 1 + 3 + .. + (2N – 1) = N2 is true.)k=1
N
For n = N+1, the left hand sum is
(2k – 1) = 1 + 3 + .. + (2N – 1) + [2(N+1) – 1]k=1
N+1
by the induction assumation this sum is N2, hence
= N2 + [2(N+1) – 1]
= N2 + 2N + 1
= (N + 1)2 Done.
From this, verify the formula will work for n = N+1.
(i.e. to show (2k – 1) = (N+1)2 )
k=1
N+1
Therefore (2k – 1) = n2 for all natural numbers n.k=1
n

I. Verify its true when n = 1.
If n = 1, we check that its true that 1 + 1 < 21.
Example B.
Verify that n + 1 < 2n for all numbers n =1, 2, 3,…
II. Assume the inequality is true for n = N, that is,
it's true that N + 1 < 2N, then verify that it has to be
true when n = N + 1, i.e (N + 1) + 1 < 2N+1.
When n = N + 1, the left hand side is
(N + 1) + 1 < 2N + 1 < 2N + 2N < 2*2N < 2N+1. Done.
by the induction assumation this is < 2N
Use induction to verify this.

Exercise A.
Write the following sums in the  notation.
(Do not find the sums). For example, using k as the index,
1 + 2 + 3 + 4 .. + N is k.
k=1
N
1. Given N, the sum 12 + 22 + 32 + 42 .. + N2
2. Given K, the sum 1(2) + 2(3) + 3(4) + 4(5) .. + K(K+1)
4. Given D, the sum 3(5/2)2 + 3(6/2)2 + 3(7/2)2 + .. + 3(D/2)2
3. Given M, the sum 3(4) + 5(6) .. + (2M – 1)(2M)
5. Given D, the sum
(5/6)2 + (6/7)2 + (7/8)2 + .. + [(2D – 1)/(2D)]2
6. Given T, the sum 2(34) + 2(36) + 2(38) +.. + 2(32T)
7. Given N, the sum
1,000(e0.03) + 1,000(e0.03)2 + 1,000(e0.03)3 .. + 1,000(e0.03)N – 1

C. Verify that each of the following formulas is true for all
positive n using mathematical induction by completing
the following steps:
a. Tell the readers that an induction argument is coming.
b. Using k as the index, show the formula holds for n = 1.
c. Assume the formula works for the case n = N – 1
and write this “true” statement in the expanded form.
Using this fact to show the statement has to be true
for the next case when n = N
3. 1 + 2 + 3 + 4 .. + n = n (n + 1)/2
2. 1 + 3 + 5 + .. + (2n – 1) = n2
4. 2 + 6 + 10 + ... + (4n – 2) = 2n2
1. 2 + 4 + 6 + .. + 2n = n(n + 1)
5. 13 + 23 + 33 + 43 .. + n3 = n2(n + 1) 2 / 4

D. Depending on the forms of the problems, the inductive
steps maybe phrased differently.
Verify each of the following formula is true for all
positive N using mathematical induction by completing
b. Show the formula holds for N = 1.
c. Assuming the formula works for the case for N – 1
and write this “true” statement in the expanded form.
Using this fact to show the statement has to be true for the
next case for N.
6. 𝑘=1
𝑁
2 𝑘 = 2N+1 – 2
7. 𝑘=1
𝑁
𝑘(𝑘 + 1) = N(N + 1)(N + 2)/3
8. 𝑘=1
𝑁.
1/[k(k + 1)] = 1 – 1/(N + 1)
9. 𝑘=1
𝑁.
2/[k(k + 2)] = 1 – 2/(N + 2)

E. Verify each of the following formula is true for all
positive N using mathematical induction by completing
b. Show the statement is true for N = 1.
c. Assuming the statement is true for the case N – 1
and write down this “true” statement.
Then use this fact to establish that the next case for N,
the statement must also be true.
1. 3N - 1 is divisible by 2.
4. 2N3 – 3N2 + N is divisible by 6.

(Answers to the odd problems) Exercise A.
k=1
N
1.  k2
k=2
M
3.  (2k – 1)(2k)
k=3
D
5.  [(2k – 1)/(2k)]2
k=2
N
7.  1,000(e0.03)k – 1
Exercise B.
1. 3 + 5 + … + [2(k – 1) +1] + [2k +1]
3. 2 + 5 + … + [3(n – 2) –1] + [3n –1]
5. 13 + 16 + … + [3(N – 2) –20] + [3(N – 1) –20]
7. – 7 – 10 – … – [2 – 3(2K– 2)] + [2 – 3(2K– 1)]
Exercise C.
1. We will use induction
I. Verify it’s true when n = 1
If n = 1 we have 2k = 2(1) = 2 = 1(1 + 1)k=1
1

II. Assume the formula works for n = N – 1
i.e., 2k = 2 + 4 + 6 + … + 2(N – 1) = (N – 1)(N)k=1
N – 1
III. Verify that the formula works for n = N
For n = N, we have 2k =  2k + 2N = (N – 1)(N) + 2Nk=1
N
k=1
N – 1
= N2 – N + 2N
= N2 + N = N(N + 1)
If n = 1 we have k = 1 = 1(1 + 1)/2k=1
1
i.e., k = 1 + 2 + … + (N – 1) = (N – 1)(N)/2k=1
N – 1
For n = N, we have  k =  k + N = (N – 1)N/2 + Nk=1
N
k=1
N – 1
= (N2 – N + 2N)/2
= (N2 + N)/2 = N(N + 1)/2

If n = 1 we have k3 = 13 = 1 = 12(1 + 1)2/4k=1
1
i.e., k3 = 13 + 23 + … + (N – 1)3 = (N – 1)2N2/4k=1
N – 1
For n = N, we have  k3 =  k3 + N3 = (N – 1)2N2/4 + N3
k=1
N
k=1
N – 1
= (N4 – 2N3 + N2 + 4N3)/4
= N2(N2 + 2N + 1)/4
= (N4 + 2N3 + N2)/4
= N2(N + 1)2/4
If n = 1 we have  k(k + 1) = 1(1 + 1) = 2 = 1(1 + 1) (1 + 2)/3k=1
1
Exercise D.

i.e.,  k (k + 1) = 2 + 6 + … + (N – 1)N = (N – 1)(N)(N + 1)/3k=1
N – 1
For n = N, we have  k(k + 1) =  k(k + 1) + N(N + 1)k=1
N
k=1
N – 1
= (N – 1)(N)(N + 1)/3 + N(N + 1)
= [(N – 1)(N)(N + 1) + 3N(N + 1)]/3
= N[(N – 1)(N + 1) + 3(N + 1)]/3
= N[N2 – 1 + 3N + 3]/3
= N[N2 3N + 2]/3 = N(N+1)(N+2)/3
9. The statements is not true, because when n = 1, we have
𝑘=1
1.
2/[k(k + 2)] = 2/[1(1+2)] = 2/3 = 1 – 2/(1 + 2)

If n = 1 we have 31 – 1 = 2, which is divisible by 2.
i.e. 3N-1 – 1 is divisible by 2.
For n = N, we have 3N – 1 = 3(3N-1 – 1) + 2
Exercise E.
divisible by 2 divisible by 2
If n = 1 we have 71 – 1 = 6, which is divisible by 6.
i.e. 7N-1 – 1 is divisible by 6.
For n = N, we have 7N – 1 = 7(7N-1 – 1) + 7

5.4 mathematical induction t

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5.4 mathematical induction t