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... :(

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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

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.

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)**

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

**P(k+1)**

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

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

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

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

Hence P(k+1) IS true, IF P(k) is true.

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?

Re: Mathematical induction: sum of odd numbers.

Re: Mathematical induction: sum of odd numbers.

$\displaystyle \begin{aligned} & S = \sum_{1 \le k \le n}\left(2k-1\right) = \sum_{1 \le n-k \le n}\left[2(n-k)-1\right] = \sum_{1-n \le -k \le 0}\left(2n-2k-1\right) \\& = \sum_{0 \le k \le n-1}\left(2n-2k-1\right) = (n-1)-(2n-2n-1)+\sum_{1 \le k \le n}\left(2n-2k-1\right). \\& \begin{aligned}\therefore ~ 2S & = 2n+\sum_{1 \le k \le n}(2k-1+2n-2k-1) = 2n+\sum_{1 \le k \le n}(2n-2) \\& = 2n+(2n-2) \sum_{1 \le k \le n}(1) = 2n+2n(n-1) = 2n^2. \\& \therefore ~ S = n^2.\end{aligned} \end{aligned}$

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

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!