{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Abstract Alg. 1

# Abstract Alg. 1 - 2 Label this equation which we want to...

This preview shows pages 1–3. Sign up to view the full content.

2. Label this equation, which we want to verify, 1 + 3 + 5 + 7 + …+ ( 2 n - 1 ) = n 2 with an asterisk (*).1 + 3 + 5 + 7 + …+ ( 2 n - 1 ) = n 2 (*). We can verify by hand calculation that (*) is correct when n is 1: When n = 1, the left side of (*) is just 1.And the right side of (*) is 1 2 , which is also 1. So the left and right side of (*) are indeed equal when n = 1. Now assume for the induction hypothesis that (*) is correct, we now have to check whether or not (*) is correct when we replace n by n+1 on each side of (*). So here is what we get to assume this is true: 1 + 3 + 5 + 7 + …+ ( 2 n - 1 ) = n 2 (the "induction hypothesis") And here is what we have to try to prove: 1 + 3 + 5 + 7 + …+ ( 2 n - 1 ) + ( 2 [ n + 1 ] - 1 ) =?? ( n +1) 2 call this (**). We know from the induction hypothesis that 1 + 3 + 5 + 7 + …+ ( 2 n - 1 ) = n 2 Using this induction hypothesis to simplify the expression 1 + 3 + 5 + 7 + …+ ( 2 n - 1 ) by just replacing this whole sum with n 2 , we can write the left side of (**) in a much simpler-looking way:1 + 3 + 5 + 7 + … + ( 2 n - 1 ) + ( 2 [ n + 1 ] - 1 ) = n 2 + ( 2 [ n + 1 ] - 1 ) because we know from the induction hypothesis that 1 + 3 + 5 + 7 + …+ ( 2 n - 1 ) = n 2 Therefore we now know 1 + 3 + 5 + 7 + …+ ( 2 n - 1 ) + ( 2 [ n + 1 ] - 1 ) = n 2 + ( 2 [ n + 1 ] - 1 ) And now notice that n 2 + ( 2 [ n + 1 ] - 1 ) = n 2 + 2 n + 2 - 1 = n 2 + 2 n + 1 = ( n +1) 2 We have now completed the proof of (**) because we showed that 1 + 3 + 5 + 7 + …+ ( 2 n - 1 ) + ( 2 [ n + 1 ] - 1 ) does in fact = ( n +1) 2 . This completes the induction step and as a result we can now assert that (*) is true for every natural number n. Theorem 2: Prove that 1 + 3 + 5 + . . . + (2n − 1) = n2 (_) for any integer n _ 1. Proof: STEP 1: For n=1 (_) is true, since 1 = 12. STEP 2: Suppose (_) is true for some n = k _ 1, that is 1 + 3 + 5 + . . . + (2k − 1) = k2. STEP 3: Prove that (_) is true for n = k + 1, that is 1 + 3 + 5 + . . . + (2k − 1) + (2k + 1) ?= (k + 1)2. We have: 1 + 3 + 5 + . . . + (2k − 1) + (2k + 1) ST.2 = k2 + (2k + 1) = (k + 1)2. _ 4. Problem: For any natural number n , n 3 + 2n is divisible by 3. This makes sense Proof:

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Basis Step: If n = 0, then n 3 + 2n = 0 3 + 2*0 = 0. So it is divisible by 3. Induction:
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 6

Abstract Alg. 1 - 2 Label this equation which we want to...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online