Question

Establish the given formula with mathematical induction:
1 + 3 + 5 + ... + (2n - 1) = n^2 for all n >= 1

Answer 1

\[1 + 3 + 5 + ... + (2n - 1) = n^2\]

Answer 2

What I tried:
\(1 + 3 + 5 + ... + (2k - 1) + (k + 2) = k^2 + k + 2\)
As I understand, I should make the right side to be a square, however, to do that I need 2 and 3k, What should I do?

Answer 3

I can rework left hand side to: \(1 + 3 + 5 + ... + k + (2k +1)\) easily.

(Oh, and to clarify - I took two numbers for induction: k and k + 2)

Answer 4

The right side should be (k+1)^2=k^2+2k+1

Answer 5

The right side should be (k + 2)^2 actually. Notice that the formula is for odd integers.

Answer 6

The 2n appears on the left side only.
Also, the next term is not k+2, it's (2(k+1)-1)=2k+1

Answer 7

But aren't the terms written as k_1 = 1; k_2 = 3; k_3 = 5? then wouldn't it be one term k and the other k + 2?

Answer 8

Best way is try not to jump steps.
Replace the last k with (k+1) and simplify.
1+3+5+...(2n-1) = n^2 ....... assumed valid for n=k, then
1+3+5+...(2k-1) = k^2........for k
Add one more term on each side, i.e. k-> k+1
1+3+5+...(2k-1) + (2(k+1)-1) = (k+1)^2
...
and continue.

Answer 9

But why isn't the next term k+2?
As far as I see, the terms on the left side are 1, 3, 5 and so on. So, why are you insisting that, if we took a term as k, then the next one would be k + 1?

Answer 10

My logic says that if the term k = 1, then the next term, which is equal to 3, isn't k + 1, but it's k + 2.

Answer 11

The left side jumps by two because there is a factor 2n. The right side is only n.
So the next sum is when n-> n+1
so 2(n-1)->2(n+1)-1 and
n^2->(n+1)^2
as I have written with k->k+1

If in doubt, use numbers,
1=1^2
1+3=2^2
1+3+5=3^2
...

Answer 12

But what you wrote with numbers is what I wrote with letters, isn't it?
first term k_1 = 1, then 2k - 1 = k^2, if k = k_1, then 2 * 1 - 1 = 1^2; 1 = 1

Answer 13

k_2 = 3 (or k_2 = k_1 + 2)

Answer 14

So, the terms advance with +2.

Answer 15

\[k_{n + 1} = k_n + 2\]

Answer 16

Yes, the terms (on the left hand side advance with +2 because the general term is (2k-1).
The sum advances by one because it's k^2.
The next sequence/sum is k->k+1.

Answer 17

How is it possible that k would be equal different values at the same time?

Answer 18

No, k->k+1.
The two in the sequence (LHS) is taken care of by the fact that the general term is (2k-1)...automatically.

Answer 19

Can we rewind and start from the given sequence?

1+3+5+...(2n-1) = n^2 ..................(1)

Answer 20

Ohh! I got it now.
Could you explain how to get the square at the RHS?

Answer 21

\[1 + 3 + 5 + ... + (2k - 1) + 2k = k^2 + k + 1\]
but k^2 + k + 1 isn't equal to (k + 1)^2

Answer 22

You were almost there, it's just a little algebraic mishaps that stopped you!
You jumped steps, that is why it does not work.
We'll increase k by 1, i.e k->k+1
1+3+5+....(2(k+1)-1) = (k+1)^2
becomes
1+3+5+....(2k-1)+(2(k+1)-1) = (k+1)^2
now continue, but be very careful with algebra!

Answer 23

Hm, thank you! :)

Answer 24

Tell me if it works out for you.
Post what you've got if you wish.

Answer 25

Tried to do it with adding k + 1 to both sides (as I understand, it should work the same) and got this:
\[1 + 3 + 5 + ... + (2k - 1) + k + 1 = k^2 + k + 1\] got everything done with LHS:
\[1 + 3 + 5 + ... + k + 2k = k^2 + k + 1\]
However, RHS should be \[k^2 + k + 1 = (k + 1)^2\], but it's not, since \[k^2 + 2k + 1 = (k + 1)^2\]

Answer 26

You jumped steps again! :(
1+3+5+...+(2k−1)+(k+2)=k2+k+2
should read
\(1+3+5+...+(2k−1)+\color{red}{2(k+1)-1}=k^2+2k+1 \)
recall
(k+1)^2=k^2+2k+1
and
2(k+1)-1=2k+2-1=2k+1

Answer 27

Ahh! Feeling so dumb right now :/ Thank you for pointing it out, understood it now :)

Answer 28

Good! Review a little on algebra and it will help a ton!!! :)

Answer 29

Might have to do so :D

Answer 30

Great!
So for the next problem, check out the algebra carefully.
Your logic is sound.! :)