Question

Hello can you help me? I don't understand somethings by this exercise from Induction.
I have load the picture.

What I can't understand is circled in red.

First why the k change I mean, why k assume number value?

In this picture the k dont change:

http://openstudy.com/users/franzmller682#/updates/54a872d1e4b054f0c3b92c0f


Assume it is true for n=n+1

In this we have 2n+2

But in the first power we have (-1)^2n+1 then 2n+1 but we have 2n+2 and that cames in the last power.

Can you make an example with numbers? I mean subtuite n with numer?

thx

Answer 1

oh I see

Answer 2

the reason we assume that the statement is true for some k, is that it is true for k=1
we can write the number 1 as k, and since we proved the statement is true for 1, we proved that it is true for at least *some* k (namely, k=1)

Answer 3

if you're not understanding please say so

Answer 4

ok

Answer 5

sorry my internet

Answer 6

we have two powers in the first are (-1)^2n+1 why 2n+1 when we have 2n+2

Answer 7

because they "stripped out" or "removed" the last two terms of the series
consider:\[\sum_0^ni=\sum_0^{n-1}i+n\]understand?

Answer 8

can you do example strippe with 2n+2?

Answer 9

I'm not sure what you mean\[\sum_0^{2n+2}i=\left[\sum_0^{2n+1}i\right]+(2n+2)=\left[\sum_0^{2n}i\right]+(2n+1)+(2n+2)\]

Answer 10

Ok this is it... this I want understand

Answer 11

Ok we have 2n+2 now it became the sum 2n+1 this step how I can do it

Answer 12

well look at a simpler case:\[\sum_0^5i=\left[\sum_0^4i\right]+5\]do you understand this completely?

Answer 13

No but it must be at step the 5 became for and I add 5

Answer 14

for=four

Answer 15

I don't think I understand, I'm sorry
I know the language barrier is difficult
You don't understand what I posted above?

Answer 16

Ok.. I dont understand it because I see only 4 and add 5 in the second. Sow it are more steps behind there?

Answer 17

\[1+2+3+4+5=\left[\sum_0^1i\right]+2+3+4+5=\left[\sum_0^2i\right]+3+4+5\\\left[\sum_0^3i\right]+4+5=\left[\sum_0^4i\right]+5=\sum_0^5i\]
Do you follow this at all?

Answer 18

yes

Answer 19

Ok but now I dont understand why k became a number value

Answer 20

\[1+2+3+4+5=\left[\sum_0^1k\right]+2+3+4+5=\left[\sum_0^2k\right]+3+4+5\\\left[\sum_0^3k\right]+4+5=\left[\sum_0^4k\right]+5=\sum_0^5k\]let's choose a more complicated summand than just k by itself.

Answer 21

\[\sum_0^{4}2k+1=\left[\sum_0^{3}2k+1\right]+2(4)+1=\left[\sum_0^{3}2k+1\right]+9\]

Answer 22

now let us instead use \(n=4\) instead of writing out the upper limit of the sum\[\sum_0^{n}2k+1=\left[\sum_0^{n-1}2k+1\right]+2n+1\]

Answer 23

it is the exact same thing as the case you understood, just with the limits written in terms of \(n\) instead of anactual number

Answer 24

please let me know if this is making any sense to you

Answer 25

ok give me time to understand please when you will I can send a massage.

Answer 26

when you want

Answer 27

sure, take your time

Answer 28

Ok I understand it but in this picture the k dont change:
http://assets.openstudy.com/updates/attachments/54a872d1e4b054f0c3b92c0f-franzmller682-1420906005020-exercisei0003.jpg

Answer 29

\[1+2+3+4+5=\left[\sum_0^1k\right]+2+3+4+5=\left[\sum_0^2k\right]+3+4+5\\=\left[\sum_0^3k\right]+4+5=\left[\sum_0^4k\right]+5=\sum_0^5k\\\text{let }n=2,\text{ we can then rewrite this as:}\]\[(2n-3)+(2n-2)+(2n-1)+2n+(2n+1)\\=\left[\sum_0^{2n-3}k\right]+(2n-2)+(2n-1)+2n+(2n+1)\\=\left[\sum_0^{2n-2}k\right]+(2n-1)+2n+(2n+1)\\=\left[\sum_0^{2n-1}k\right]+2n+(2n+1)=\left[\sum_0^{2n}k\right]+(2n+1)=\sum_0^{2n+1}k\]Your problem simply runsthis processbackwards.

Answer 30

runs this process backwards*

Answer 31

Basically, sigma notation is used to represent adding a sequence of numbers. In your problem, you can remove terms from the sequence, as long as you add what you removed outside of the sigma expression.

Answer 32

\[\sum_0^{4}2k+1=\left[\sum_0^{3}2k+1\right]+2(4)+1=\left[\sum_0^{3}2k+1\right]+9\]