
In Problems 1-22, use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers .
is divisible by 2.

To prove: The given statement is divisible by 2 is true for all natural numbers using the Principle of Mathematical Induction.
Answer to Problem 21AYU
As the statement is true for the natural number terms, hence the statement is true for all natural numbers.
Explanation of Solution
Given:
Statements says the series is divisible by 2 is true for all natural number.
Formula used:
The Principle of Mathematical Induction
Suppose that the following two conditions are satisfied with regard to a statement about natural numbers:
CONDITION I: The statement is true for the natural number 1.
CONDITION II: If the statement is true for some natural number , it is also true for the next natural number . Then the statement is true for all natural numbers.
Proof:
Consider the statement
is divisible by 2 -----(1)
Step 1: Show that statement (1) is true for the natural number .
That is is divisible by 2. Hence the statement is true for the natural number .
Step 2: Assume that the statement is true for some natural number .
That is is divisible by 2 -----(2)
Step 3: Prove that the statement is true for the next natural number .
That is, to prove that is divisible by 2
Consider
From equation (2), first term in the above equation is divisible by 2 and second term is a multiple of 2.
Hence is divisible by 2.
As the statement is true for the natural number terms, hence the statement is true for all natural numbers.
Chapter 12 Solutions
Precalculus Enhanced with Graphing Utilities
Additional Math Textbook Solutions
College Algebra (7th Edition)
Elementary Statistics: Picturing the World (7th Edition)
Thinking Mathematically (6th Edition)
Pre-Algebra Student Edition
A First Course in Probability (10th Edition)
Intro Stats, Books a la Carte Edition (5th Edition)
- For the system consisting of the lines: and 71 = (-8,5,6) + t(4, −5,3) 72 = (0, −24,9) + u(−1, 6, −3) a) State whether the two lines are parallel or not and justify your answer. b) Find the point of intersection, if possible, and classify the system based on the number of points of intersection and how the lines are related. Show a complete solution process.arrow_forward3. [-/2 Points] DETAILS MY NOTES SESSCALCET2 7.4.013. Find the exact length of the curve. y = In(sec x), 0 ≤ x ≤ π/4arrow_forwardH.w WI M Wz A Sindax Sind dy max Утах at 0.75m from A w=6KN/M L=2 W2=9 KN/m P= 10 KN B Make the solution handwritten and not artificial intelligence because I will give a bad rating if you solve it with artificial intelligencearrow_forward
- Solve by DrWz WI P L B dy Sind Ⓡ de max ⑦Ymax dx Solve by Dr ③Yat 0.75m from A w=6KN/M L=2 W2=9 kN/m P= 10 KN Solve By Drarrow_forwardHow to find the radius of convergence for the series in the image below? I'm stuck on how to isolate the x in the interval of convergence.arrow_forwardDetermine the exact signed area between the curve g(x): x-axis on the interval [0,1]. = tan2/5 secx dx andarrow_forward
- Calculus: Early TranscendentalsCalculusISBN:9781285741550Author:James StewartPublisher:Cengage LearningThomas' Calculus (14th Edition)CalculusISBN:9780134438986Author:Joel R. Hass, Christopher E. Heil, Maurice D. WeirPublisher:PEARSONCalculus: Early Transcendentals (3rd Edition)CalculusISBN:9780134763644Author:William L. Briggs, Lyle Cochran, Bernard Gillett, Eric SchulzPublisher:PEARSON
- Calculus: Early TranscendentalsCalculusISBN:9781319050740Author:Jon Rogawski, Colin Adams, Robert FranzosaPublisher:W. H. FreemanCalculus: Early Transcendental FunctionsCalculusISBN:9781337552516Author:Ron Larson, Bruce H. EdwardsPublisher:Cengage Learning





