# Homework Help: Solve this mathematical induction

1. Aug 27, 2014

### mxam

Can you help me with this exercise?

$1^{1}$+$2^{2}$+$3^{3}$+$4^{4}$+...+$n^{n}$ = $n^{n+1}$

Thanks!

PD. I was trying to solve, and i have this:

$1^{1}$=$1^{1+1}$ =
1 = 1

a) $k^{k+1}$
b) $k+1^{k+1}$ = $k+1^{(k+1)+1}$

a in b) $k^{k+1}$ + $k+1^{k+1}$ = $k+1^{(k+1)+1}$

($k)^{k}$($k)^{1}$+($k+1)^{k}$($k+1)^{1}$=($k+1)^{k}$($k+1)^{1}$($k+1)^{1}$

I´m lost in this step . . . Thanks again!

Last edited: Aug 27, 2014
2. Aug 27, 2014

### dextercioby

Where did you pick the exercise from? It looks wrong.

3. Aug 27, 2014

### mxam

The Algebra Teacher gives the exercise, but i dont know, maybe don´t have a solution . . . ?

4. Aug 27, 2014

### Staff: Mentor

The series and its sum don't seem correct.

Can you check your book again?

1+2^2 =/= 2^(2+1)

1+4 = 5 but 2^(2+1) = 8

5. Aug 27, 2014

### Staff: Mentor

There are many series that look vaguely similar like

1 + 2 + 3 + ... = (n*(n+1))/2

6. Aug 27, 2014

### mxam

A attach the picture of the book. . .

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7. Aug 27, 2014

### Staff: Mentor

Can you say there is no solution and prove that by example?

8. Aug 27, 2014

### mxam

The teacher said that exist a solution, but with the opinions and the examples i think that he´s lying. . . =o!

9. Aug 27, 2014

### Staff: Mentor

Maybe strategically misdirecting you in order to have you discover this alternative...

You could PM a PF Mentor to see if they can add some new insight...

10. Aug 27, 2014

### gopher_p

I am curious. What are the instructions for the exercise? You did not include them in the original post, and your picture does not show them either.

11. Aug 28, 2014

### mxam

Check the Attachment, the exercise says: "Demuestre por Induccion Matematica las siguientes preposiciones dadas" or in English, "Demonstrate by Mathematical Induction the following prepositions" . . . I think is a exercise that dont have a real solution, maybe the teacher´s proving us, . . . But I´ll need a good explanation for this, =)

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12. Aug 28, 2014

### vanhees71

The formula is obviously wrong.

Much more interesting is the question, how to figure out such sums. One clever trick is to use an easier sum to derive the one you want to know. Here the obvious idea is to use the generating function
$$g(x)=\sum_{k=0}^n \exp(k x).$$
This is a geometric series. You should prove, e.g., by induction, that it gives
$$g(x)=\frac{\exp[(n+1) x)-1}{\exp x-1}.$$
Then you can get other sums by taking derivatives of this, e.g.,
$$g'(x)=\sum_{k=0}^{n} k \exp(k x) \; \Rightarrow \; g'(0)=\sum_{k=0}^n k,$$
and so on. There's, of course a subtlety with taking $x=0$, which you also should discuss.

13. Aug 28, 2014

### HallsofIvy

Yes, this is true for n= 1. But surely you saw that when n= 2, the left side is $1^1+ 2^2= 1+ 4= 5$ while the right side is $2^3= 8$.

14. Aug 28, 2014

### mxam

I´ve 1 curious question:

If i wanna know what´s the correct serie for $n^{n+1}$, what´s the the procedure to follow?

15. Aug 29, 2014

### dirk_mec1

Use a Taylor series.

16. Aug 29, 2014

### Staff: Mentor

Doing the first nine iterations:

Code (Text):

n=1:    1               sum/psum = 1
n=2:    5               sum/psum = 5
n=3:    32              sum/psum = 6
n=4:    288             sum/psum = 9
n=5:    3413                sum/psum = 11
n=6:    50069               sum/psum = 14
n=7:    873612              sum/psum = 17
n=8:    17650828        sum/psum = 20
n=9:    405071317       sum/psum = 22

Processing IDE code used:

Code (Text):

void setup() {

int nterms = 10;
int psum=1, sum=0;

for(int n=1; n<nterms; n++) {
sum+=Math.pow(n,n);
int iratio = sum/psum;
println("n="+n+":\t"+sum+"\t\tsum/psum = "+iratio);
psum=sum;
}

}