Step 1: Solving 1 Base Step For n21. 029<1 Let for => a ≤1, which is true. n=k, the statement is true. HaER, NKEN, [oca<1] = a* ≤1. We prove a that it is true for n=k+1 also. k+ 1 K. a. = ≤ 1 a K+1 ≤ 1 к a We have, K and a≤1. product So, their is also ≤1 So, by principle of mathematical induction, HaER, In EN, [o ca<] a" ≤1. Step 2: Solving 2 2 For n = 3 n! = 3! = 6 nn = 3³ = 27. 6 < 27. it is true. Let the statement is true 1 Now, we will prove for nak+1. for nak ie, K k! 2 >n! 2n!
Step 1: Solving 1 Base Step For n21. 029<1 Let for => a ≤1, which is true. n=k, the statement is true. HaER, NKEN, [oca<1] = a* ≤1. We prove a that it is true for n=k+1 also. k+ 1 K. a. = ≤ 1 a K+1 ≤ 1 к a We have, K and a≤1. product So, their is also ≤1 So, by principle of mathematical induction, HaER, In EN, [o ca<] a" ≤1. Step 2: Solving 2 2 For n = 3 n! = 3! = 6 nn = 3³ = 27. 6 < 27. it is true. Let the statement is true 1 Now, we will prove for nak+1. for nak ie, K k! 2 >n! 2n!
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Please help me with these question. I am having trouble understanding what to do. Please tell me if the following question below and there answers to them are correct.
Image 1: The 2 Questions 1 &2
Image 2: The Answers to the question (top part of the image is the answer to question #1, while the botton is answer to question #2)
Thank you
![Step 1: Solving 1
Base Step
For
n21.
029<1
Let
for
=> a ≤1, which is true.
n=k,
the statement is true.
HaER, NKEN, [oca<1] = a* ≤1.
We
prove
a
that it is true for n=k+1 also.
k+ 1 K. a.
=
≤ 1
a
K+1 ≤ 1
к
a
We have,
K
and a≤1.
product
So, their
is
also ≤1
So, by principle of mathematical
induction, HaER, In EN,
[o ca<] a" ≤1.
Step 2: Solving 2
2
For
n = 3
n! = 3! = 6
nn = 3³ = 27.
6 < 27.
it is true.
Let
the
statement is true
1
Now, we
will
prove for nak+1.
for nak ie,
K
k! <K,
LHS = (k+1)!
=
(K+D)K!
< (k+1) K
< (K+D(k+1)*
501
1
(since, k<k+1
<(x+1)").
(K+D!
(K+DK+1 = RHS.
(k+1)+1.
So, K
Hence, by induction, the N, [n> 2 >n! <n^].
(Proved).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8e1fe2dc-a274-4b26-849f-c30d3fc856c1%2Ff421460f-d924-467c-aa87-4af4f3ba5781%2Fqonj7g8_processed.png&w=3840&q=75)
Transcribed Image Text:Step 1: Solving 1
Base Step
For
n21.
029<1
Let
for
=> a ≤1, which is true.
n=k,
the statement is true.
HaER, NKEN, [oca<1] = a* ≤1.
We
prove
a
that it is true for n=k+1 also.
k+ 1 K. a.
=
≤ 1
a
K+1 ≤ 1
к
a
We have,
K
and a≤1.
product
So, their
is
also ≤1
So, by principle of mathematical
induction, HaER, In EN,
[o ca<] a" ≤1.
Step 2: Solving 2
2
For
n = 3
n! = 3! = 6
nn = 3³ = 27.
6 < 27.
it is true.
Let
the
statement is true
1
Now, we
will
prove for nak+1.
for nak ie,
K
k! <K,
LHS = (k+1)!
=
(K+D)K!
< (k+1) K
< (K+D(k+1)*
501
1
(since, k<k+1
<(x+1)").
(K+D!
(K+DK+1 = RHS.
(k+1)+1.
So, K
Hence, by induction, the N, [n> 2 >n! <n^].
(Proved).
![1. Prove that
Va ER, Vn Є N, [0 < a < 1] ⇒ an ≤1
using mathematical induction. Justify every step.)
2. Prove that
Vn EN, [n>2n!<n"]
using mathematical induction. Justify every step.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8e1fe2dc-a274-4b26-849f-c30d3fc856c1%2Ff421460f-d924-467c-aa87-4af4f3ba5781%2Fpjh2t31_processed.jpeg&w=3840&q=75)
Transcribed Image Text:1. Prove that
Va ER, Vn Є N, [0 < a < 1] ⇒ an ≤1
using mathematical induction. Justify every step.)
2. Prove that
Vn EN, [n>2n!<n"]
using mathematical induction. Justify every step.)
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