Mathematica Proving Math Induction for S1-S3: 1+3+5+...+(2n-1)=n^2

[Blank101]

Prove that S1, S2, S3 are true statements
1+3+5+...+(2n-1)=n^2


S1=1= (2(1)-1) = 1^2 True
S2=1+3 = (2(3)-1) = 5 which cannot= to the sum of our first 2 integers, which will make it false!
S3=1+3+5 = (2(5)-1) = 3^2 True

The problem is with S2 the book gave me an answer of 4=4 which is 2^2!
It also shows a different formula (n-1) = n^2

In my understanding of The mathematical induction a formula is usually given to prove an x number of integers, all those integers being proven true does not mean will be the same to all integers from the sequence. Thats when K+1 substitution comes in.

Can someone help me i know is a simple mistake but i can't just see why s2 is coming that way!

Thanx

 

[rock.freak667]

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

the (2n-1) gives the nth term.
So the second term is 2(2)-1=3 (as seen in the series)

and so S2=1+3=4 (LHS)
and S2=2^2 (RHS)

LHS=RHS so it's true

 

[Defennder]

That's an odd "proof" of induction. You've verified it for S1, S2, S3, but these are just particular instances of the problem you're purportedly trying to prove by induction, not the general "k+1" case.

 

[HallsofIvy]

That's an odd "proof" of induction. You've verified it for S1, S2, S3, but these are just particular instances of the problem you're purportedly trying to prove by induction, not the general "k+1" case.

I don't believe he is claiming that as a "proof". He was asserting that the statement was not true. It is true, of course, he simply did not understand what the formula said:

S1=1= (2(1)-1) = 1^2 True
S2=1+3 = (2(3)-1) = 5 which cannot= to the sum of our first 2 integers, which will make it false!
S3=1+3+5 = (2(5)-1) = 3^2 True

No, the statement does NOT say 1+ 3= 2(3)- 1, it says 1+ 3= 1+ (2(2)-1)= 2^2. It is the last integer that is "2n- 1", not the sum.

In fact, your statement about S3 is incorrect: 1+ 3+ 5= 1+ [2(2)- 1]+ [2(3)-1]= 3^2 is what it says.