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?

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.

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!