Question

Help with Mathematical Induction.

Answer 1

My goal is to prove that the following statement is true for all positive integers of n:
\[4\times6+5\times7+6\times8+...+4n(4n +2)=\frac{ 4(4n+1)(8n+7) }{ 6 }\]

Answer 2

did you check the case for \(n=1\) ?

Answer 3

or if it is false show why. I am not sure how I am suppossed to do this though because on the lesson it is showing someone solving a problem similar.

Answer 4

you get to assume it is true for \(n=k\) in other words you get to assume
\[4\times6+5\times7+6\times8+...+4k(4k +2)=\frac{ 4(4k+1)(8k+7) }{ 6 }\]

Answer 5

before we begin, have you done a proof by induction before? or is this an on line class and you have only seen on example done via video or something?

Answer 6

It is online I just watch videos on how to do it and in the videos they are like solving the problem for n?

Answer 7

not to be discouraging, but this is a technique of proof that is hard to understand via a video. it is not at all like solving for \(n\) .

Answer 8

and this is a really bad example to start with, because the algebra is going to be rather complicated, and that is after you understand what the structure of the proof looks like.

Answer 9

I just got finished learning about arithmetic and geometric sequences.

Answer 10

try looking at some examples from patrick at just math tutoring, like this one
www.youtube.com/watch?v=IFqna5F0kW8

Answer 11

S(k+1) = 1 + 3 + 5 + ... + (2k – 1) + (2(k+1) – 1);
S(k+1) = 1 + 3 + 5 + ... + (2k – 1) + (2k +2 – 1); distributed the 2

= 1 + 3 + 5 + ... + (2k – 1) + (2k +1); combined like terms


= equals k2 + (2k +1) ; based on the given Sk formula
= k2 + 2k + 1
= (k + 1)(k + 1); factor
= (k + 1)2
= S(k+1)

This is one of their examples.

Answer 12

it is going to take a while to understand how to do this, try simple examples first
for example see if you can understand how to prove
\[1+2+3+...+n=\frac{n(n+1)}{2}\]

Answer 13

the algebra in your example is going to be to show that if
\[4\times6+5\times7+6\times8+...+4k(4k +2)=\frac{ 4(4k+1)(8k+7) }{ 6 }\] then
\[4\times6+5\times7+6\times8+...+4k(4k +2)+4(k+1)(4(k+1)+2)\]
\[=\frac{ 4(4(k+1)+1)(8(k+1)+7) }{ 6 }\]

Answer 14

start with an easy example first, it is really unfair for them to ask you to do this without making sure you understand the principle of mathematical induction first

Answer 15

Alright thank you for that video I will go over it and try to figure this out.