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Question 1173524: Fill in the blanks indicated by the question marks to complete the proofs of each, mathematical statement. (in mathematical induction)
a. prove: 8n-3n is divisible by 5
proof: i/we verify the statement for n=1, i.e (1)_____ -3^1, divisible 5
II. assume that the statement is there for some integer n=k
(2)_____ is divisible by 5
III. we need to show that the statement is also true for the next integers n=k+1, i.e. 8k+1=3k+1 is divisible by 5.
consider: 8k+1-3k+1=8k*8-3k*3
=8(8k-3k)- (3)_____ but 8(8k-3k) is divisible by 5 by (4)_____ and the second term (5)_____ is obviously divisible by 5 because of the factor (6)_____.
Thanks for your help.
Answer by math_helper(2461)   (Show Source):
You can put this solution on YOUR website!
They are illustrating proof by induction
(1) 8-3 is divisible by 5 (should not have ^1)
(2) 8k-3k is divisible by 5 (this is the hypothesis)
(3) Let n=k+1; then...
8(k+1) - 3(k+1)
= 8k+8 - (3k+3)
= 8k+8 -3k-3
= (8k-3k)+5
By hypothesis, 8k-3k is divisible by 5 (that was the n=k case)
so 8k-3k+5 is also divisible by 5 (if F is divisible by 5, so is F+5)
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This part:
"III. we need to show that the statement is also true for the next integers n=k+1, i.e. 8k+1=3k+1 is divisible by 5.
consider: 8k+1-3k+1=8k*8-3k*3"
should be:
"III. we need to show that the statement is also true for the next integers n=k+1, i.e. 8(k+1)-3(k+1) is divisible by 5.
Consider: 8(k+1)-3(k+1) = 8k+8-3k-3"
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While it seems they are trying to show how proof by induction works, note the original problem "show 8n-3n is divisible by 5" is much more easily proven directly: 8n-3n = 5n and 5*(any integer) is obviously divisible by 5.
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IMO, a better problem is to show 1+2+...+n = (1/2)(n+1)(n) by induction, which is also easily shown by direct proof but not quite as obvious.
Prove S = 1+2+3+...+n = (1/2)(n+1)(n) by induction
n=1: S=1 and (1/2)(1+1)(1) = 2/2 = 1 (holds for n=1, this is the base case)
Assume true for n=k (*) (the hypothesis)
Let n=k+1: S = [ 1+2+...+k ] + (k+1) = [ (1/2)(k+1)(k) ] + (k+1)
where we applied the hypothesis (*) for bracketed [ ] items.
Factor out k+1: S = (k+1) ((1/2)(k) + 1)
since (1/2)k + 1 = k/2 + 2/2 = (k+2)/2 we end up with
S = (k+1)(k+2)/2 = (1/2)(k+1)(k+2)
( if you write this in terms of n you have S = (1/2)(n)(n+1) )
thus we have shown the hypothesis holds true for n=k+1 and the proof is complete.
...and if you wish, by direct proof:
write the su S = 1 + 2 + 3 +...+ (n-1) + n
also write it S = n +(n-1)+(n-2) +... + 2 + 1
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add both eqns 2S = (n+1)+(n+1)+(n+1)+...+(n+1)+(n+1)
The RHS has n terms of n+1, that's just n*(n+1):
2S = n*(n+1)
and finally S = (1/2)*n*(n+1)
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