Question

Prove using mathematical induction that 1^3+2^3+...+n^3=(n^2(n+1)^2)/4

Answer 1

You must do it for base case first.

Answer 2

Then after that:
\[\text{ Assume } 1^3+2^3+ \cdots +k^3=\frac{k^2(k+1)^2}{4} \text{ holds for some integer } \\ k>0\]
\[\text{ And show } 1^3+2^3+\cdots +(k+1)^3 =\frac{(k+1)^2(k+2)^2}{4} \text{ holds }\]

Answer 3

hint:
\[1^3+2^3+\cdots +(k+1)^3 \\ =1^3+2^3+ \cdots +k^3 +(k+1)^3 \\ =\frac{k^2(k+1)^2}{4}+(k+1)^3 \]

Answer 4

you must use algebra to show that equals (k+1)^2(k+2)^2/4

Answer 5

david do not make it harder than it needs to be
do not multiply all of that out

Answer 6

Let me know when and if you need a hint

Answer 7

I just don't understand when I reached Part Three of Induction Proofs "Prove true for k+1"

Answer 8

You are saying you don't understand what it means?

Answer 9

I do. I just don't know how to get the final answer.

Answer 10

Oh first step take your second term and multiply it by 4/4

Answer 11

\[=\frac{k^2(k+1)^2}{4}+\frac{4(k+1)^3}{4}\]

Answer 12

Doing this will allow us to write as one fraction

Answer 13

\[=\frac{k^2(k+1)^2+4(k+1)^3}{4}\]

Answer 14

See. That is what I don't understand.

Answer 15

the combining fractions part?

Answer 16

How about the first part \[1^3+2^3+...+n^3=\]

Answer 17

How does this work out?

Answer 18

Are we talking about the base case?
showing it is true for n=1?

Answer 19

Step 1: Show it is true for the base case. The base case for this one is showing it is true for n=1:
\[1^3 =1 =\frac{1^2(1+1)^2}{4} \text{ holds } \]

Answer 20

Base case? I don't think so. What I mean is how does it expand? For example:\[1+2+3+...+k= \]
becomes
\[1+2+3+k+ k+1\]
This is what I mean. How does
\[1^3+2^3+...+n^3 \]
become "expanded"??

Answer 21

\[\text{ Assume } 1^3+2^3+ \cdots +k^3=\frac{k^2(k+1)^2}{4} \text{ holds for some integer } \\ k>0
\text{ And show } 1^3+2^3+\cdots +(k+1)^3 =\frac{(k+1)^2(k+2)^2}{4} \text{ holds }\]
That thing I assumed is true (first line of this post) is an equation I found just from replacing n with k
The thing we want to show using the thing we assume is true is an equation I found from just replacing the n's with (k+1)'s

Answer 22

@amistre64 or @hero

Answer 23

I'm calling others here because I don't understand. I'm sorry.

Answer 24

I don't understand either. I hate Mathematical Inductions.

Answer 25

Like I mean I don't understand your question is all.

Answer 26

proof by induction is all about form .... since know a certain form is true we must manipulate the k+1th element into the 'true form'

Answer 27

Like this:
\[1+2+3+...=\frac{ n(n+1) }{ 2 }\]
\[1+2+3+...+k+k+1= \frac{ (k+1)(k+1+1) }{ 2 }\]\[\frac{ k(k+1) }{ 2} + k+1 = \frac{ (k+1)(k+2) }{ 2 }\]
\[\frac{ k(k+1)+2(k+1) }{ 2 } = \frac{ (k+1)(k+2) }{ 2}\]

See how it's "expanding" after the first line Freckles???

Answer 28

after the basis step, which proves its true for a specific value, we want to make it general again ... let n=k

1^3+2^3+...+k^3 = (k^2(k+1)^2)/4 <= it is this FORM that we know is good

if we let the left side be named, say P(k) then

P(k) = (k^2(k+1)^2)/4

add then k+1th term to each side ..

P(k) + (k+1)^3 = (k^2(k+1)^2)/4 + (k+1)

now the leftside is just the P(k+1) term, and we can focus on the right side manipulating it into the 'good' form to show that k has turned into k+1

Answer 29

forgot my ^3 on the right side :)

Answer 30

Ok. I kind of starting to get it.

Answer 31

maybe if I do an example for you or something...
Let's show \[1^2+2^2+3^2 + \cdots + n^2=\frac{n(n+1)(2n+1)}{6} \]
The first case is showing it is true for the base case.
The base case being for n=1 in this case
So we have \[1^2 \text{ on left hand side } \text{ and } \frac{1(1+1)(2(1)+1)}{6}=\frac{1(2)(3)}{6} \\ \text{ both sides are equal to 1 } \text{ this shows the base case }\]
Now we have an if then statement
\[\text{ If } 1^2+2^2+\cdots + k^2=\frac{k(k+1)(2k+1)}{6} \\ \text{ for some integer } k \ge 1 \text{ , then } \\ 1^2+2^2+\cdots +(k+1)^2=\frac{(k+1)(k+2)(2(k+1)+1)}{6} \text{ holds }\]
So in if then statements we get to suppose the if part and show the then part...
\[1^2+2^2+\cdots +(k+1)^2 \\ \text{ we know k comes before (k+1) } \\ \text{ so we can write } \\ 1^2+2^2+ \cdots +k^2+(k+1)^2 \\ \text{ but by our induction hypothesis we can write } 1^2+2^2+ \cdots + k^2 \\ \text{ as } \frac{k(k+1)(2k+1)}{6} \\ \text{ so we have } 1^2+2^2 + \cdots +k^2+ (k+1)^2 \text{ can be written as } \\ \frac{k(k+1)(2k+1)}{6} +(k+1)^2 \\ \text{ Remember we are trying to show this is equal to the one fraction } \\ \frac{(k+1)(2k+1)(2k+3)}{6} \\ \text{ so let's look at } \\ \frac{k(k+1)(2k+1)}{6} +(k+1)^2 \text{ and think ... } \\ \text{ well we need one fraction so ... let's combine fractions } \\ \frac{k(k+1)(2k+1)}{6} +\frac{6(k+1)^2}{6} \\ \frac{k(k+1)(2k+1)+6(k+1)^2}{6} \\ \text{ now do some factoring } \frac{(k+1)[k(2k+1)+6(k+1)}{6} \\ \frac{(k+1)[2k^2+k+6k+6]}{6} \\ \frac{(k+1)(2k^2+7k+6)}{6} \\ \frac{(k+1)(2k^2+3k+4k+6)}{6} \\ \frac{(k+1)[k(2k+3)+2(2k+3)]}{6} \\ \frac{(k+1)(2k+3)(k+2)}{6} \\ \text{ which is what we wanted to show } \\ \text{ so therefore we can conclude } \\ 1^2+2^2+\cdots +n^2=\frac{n(n+1)(2n+1)}{6} \text{ holds for all integer } n \ge 1. \]

Answer 32

the thing is, if we knows it true for some k (we know its true for the base case)

and can show by 'form' that the (k+1)th term fits the same format, then we have shown that if we simply replace k with k+1 that the setup remains consistent ... nothing is altered, and therefore we can say its true for the next term, and the next term, and the next ..

Answer 33

term being defined as an integer that is

Answer 34

1+2+3+...+k doesn't become 1+2+3+...+k+(k+1)

Answer 35

it does if you add the next term :) but other than that its immaterial what the left side looks like

Answer 36

saying 1+2+3+...+k becomes 1+2+3+...+k+(k+1)
feels like someone is saying these are equal so I just didn't like the word becomes

Answer 37

The statement is P(n)
and we want to show P(n) is true for all positive integer n
So show P(1) is true
assume P(k) is true
show P(k+1) is true

Answer 38

'becomes' has a connotation of change, of morphing, of changing by default ... not equality or sameness.

Answer 39

tomato tomato tho lol

Answer 40

Ok ok like the spice girls say 2 becomes 1

Answer 41

2 becomes 1 by subtracting 1 from 2

Answer 42

or there some other ways 2 can become 1

Answer 43

But I don't know if I should rely on Spice Girls for my mathematical meanings

Answer 44

she becomes pregnant infers something has changed :)

Answer 45

david, any thoughts brewing?

Answer 46

Well my math teacher taught us that k becomes k+k+1

Answer 47

Why must Mathematical Inductions be taught anyways?? I don't see it being useful in life anyways.

Answer 48

Well I know it is useful in some computer science-y things and it definitely useful for proving things... It also exercises your algebra skills.

Answer 49

Or so I heard it is useful in computer science

Answer 50

I'm more into Biology, that's why.

Answer 51

http://www.cs.sfu.ca/~bbart/225/induction1-4.html

Answer 52

This says why it is useful in computer science
But I have no clue how it will help you in bio

Answer 53

Besides to help strengthen those critical thinking skills and algebra skills

Answer 54

Exactly. This proves my point why I don't find this useful at all. I suck in math.

Answer 55

I think there is math in bio if I'm not mistaken

Answer 56

I definitely know statistics is important to every science

Answer 57

There is math in bio but it's not the hard type of math, just some hard math now and then. But never have I seen mathematical inductions being used in bio