Mathematical induction: sum of odd numbers.

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June 25th 2011, 07:55 AM
rcs

Mathematical induction: sum of odd numbers.

by Set Theory PMI

Prove: for all Integers n greater than or = to 1.

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

My initial solution but stuck as i went along the way... :(

------------------------------------------------------------------------------------------
i. when n = 1, then it is true
ii. if P (k) is true then P( k +1) is true...

im stuck then here please help i dont know what to do here nex. thanks a lot

1 + 3 + 5 +, ...(2n - 1) =n^2
then it is true
June 25th 2011, 08:03 AM
Prove It

Re: the principle of mathematical induction... can anybody help me?

Base step: When n = 1, you have

LHS = 1

RHS = 1^2 = 1 = LHS.

Base step is proven.

Inductive step: Assume P(k) is true, i.e. that 1 + 3 + 5 + ... + (2k-1) = k^2. Using this you need to show that P(k+1) is true.

For P(k + 1) you are trying to show 1 + 3 + 5 + ... + (2k-1) + (2k+1) = (k + 1)^2.

LHS = 1 + 3 + 5 + ... + (2k - 1) + (2k + 1)

= k^2 + 2k + 1 since 1 + 3 + 5 + ... + (2k - 1) = k^2

= k^2 + k + k + 1

= k(k + 1) + 1(k + 1)

= (k + 1)(k + 1)

= (k + 1)^2

= RHS.

So the Inductive step is proven.
June 25th 2011, 10:45 AM
Archie Meade

Re: the principle of mathematical induction... can anybody help me?

Quote:

Originally Posted by rcs

by Set Theory PMI

Prove: for all Integers n greater than or = to 1.

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

My initial solution but stuck as i went along the way... :(

------------------------------------------------------------------------------------------
i. when n = 1, then it is true
ii. if P (k) is true then P( k +1) is true...

im stuck then here please help i dont know what to do here nex. thanks a lot

1 + 3 + 5 +, ...(2n - 1) =n^2
then it is true

If you are stuck on the induction step,
then you might be wondering what you need to prove.

We want to show that the formula being true for "any" n
causes the formula to be true for the "next" n.

Let these successive values of n be termed "k" and "k+1".

Hence we have the following propositions

P(k)

$1+3+5+....+(2k-1)=k^2$

P(k+1)

$1+3+5+....+[2(k+1)-1]=(k+1)^2\;\;?$

We aim to show that IF P(k) is true, THEN P(k+1) is also true.

If P(k) is true, then P(k+1) becomes

$\left(1+3+5+.....+2k-1\right)+2k+1=k^2+2k+1\;\;?$

$\Rightarrow\ k^2+2k+1=k^2+2k+1$

Hence P(k+1) IS true, IF P(k) is true.
June 26th 2011, 06:43 AM
rcs

Thank you

thank you so much God BLess You are truly amazing.
June 30th 2011, 08:56 PM
rcs

RHS and LHS meant?

what does RHS and LHS meant for sir Prove It?

thanks
June 30th 2011, 09:29 PM
Prove It

Re: RHS and LHS meant?

Quote:

Originally Posted by rcs

what does RHS and LHS meant for sir Prove It?

thanks

Left Hand Side and Right Hand Side. This is standard notation.
July 1st 2011, 02:13 AM
Also sprach Zarathustra

Re: Mathematical induction: sum of odd numbers.

Quote:

Originally Posted by rcs

by Set Theory PMI

Prove: for all Integers n greater than or = to 1.

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

My initial solution but stuck as i went along the way... :(

------------------------------------------------------------------------------------------
i. when n = 1, then it is true
ii. if P (k) is true then P( k +1) is true...

im stuck then here please help i dont know what to do here nex. thanks a lot

1 + 3 + 5 +, ...(2n - 1) =n^2
then it is true

Without induction:

1 + 3 + 5 + 7 + 9 +... (2n-9)+(2n-7)+(2n-5)+(2n-3)+(2n - 1 )

Observe that:

1+(2n-1)=2n

1+(2n-1)=2n

3+(2n-3)=2n

5+(2n-5)=2n

7+(2n-7)=2n

.
.
.

The sum of all above is:

2n+2n+2n+2n+....

but how many times 2n appears?
July 1st 2011, 03:28 PM
rcs

Re: Mathematical induction: sum of odd numbers.

it is infinitely many
July 1st 2011, 04:43 PM
TheCoffeeMachine

Re: Mathematical induction: sum of odd numbers.

$\begin{aligned} & S = \sum_{1 \le k \le n}\left(2k-1\right) = \sum_{1 \le n-k \le n}\left[2(n-k)-1\right] = \sum_{1-n \le -k \le 0}\left(2n-2k-1\right) \\& = \sum_{0 \le k \le n-1}\left(2n-2k-1\right) = (n-1)-(2n-2n-1)+\sum_{1 \le k \le n}\left(2n-2k-1\right). \\& \begin{aligned}\therefore ~ 2S & = 2n+\sum_{1 \le k \le n}(2k-1+2n-2k-1) = 2n+\sum_{1 \le k \le n}(2n-2) \\& = 2n+(2n-2) \sum_{1 \le k \le n}(1) = 2n+2n(n-1) = 2n^2. \\& \therefore ~ S = n^2.\end{aligned} \end{aligned}$
July 1st 2011, 09:59 PM
Also sprach Zarathustra

Re: Mathematical induction: sum of odd numbers.

Quote:

Originally Posted by rcs

it is infinitely many

No! At the original sum there are a finite number of components so if we arrange them in pairs, we'll still have left with a finite number of components.

Try it with n=7, the sum of seven odd numbers:

1 + 3 + 5 + 7 + 9 + 11 + 13

1+13=14
3+11=14
5+9=14
and 7 left alone.

Try develop this theory of arithmetic summation by your own...

Also read here:

How To Be A Little Gauss
July 6th 2011, 08:45 AM
scruz10

Re: Mathematical induction: sum of odd numbers.

Quote:

Originally Posted by rcs

by Set Theory PMI

Prove: for all Integers n greater than or = to 1.

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

My initial solution but stuck as i went along the way... :(

------------------------------------------------------------------------------------------
i. when n = 1, then it is true
ii. if P (k) is true then P( k +1) is true...

im stuck then here please help i dont know what to do here nex. thanks a lot

1 + 3 + 5 +, ...(2n - 1) =n^2
then it is true

check if it is true for n=1
1 = 1^2
1 = 1

Assume it is true for n=k
1 + 3 + 5 + ....(2k - 1) = k^2
Check if it is true for n= k + 1
1 + 3 + 5 + ....(2k -1) + (2(k+1) - 1) = (k+1)^2
You can replace 1 + 3 +5 +....(2k-1) with k^2
k^2 + (2(k+1) - 1) = (k+1)^2
k^2 + (2k +2 - 1) = (k+1)^2
k^2 + 2k + 1 = (k+1)^2
(k+1)(k+1) = (k+1)^2
(k+1)^2 = (k+1)^2

Proven!