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 simplerlooking 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:
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View Full DocumentBasis Step:
If n = 0, then n
3
+ 2n = 0
3
+ 2*0 = 0. So it is divisible by 3.
Induction:
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 Spring '11
 Zhang
 Algebra, Mathematical Induction, Natural number, Prime number

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