document.write( "Question 1173524: Fill in the blanks indicated by the question marks to complete the proofs of each, mathematical statement. (in mathematical induction)\r
\n" ); document.write( "\n" ); document.write( "a. prove: 8n-3n is divisible by 5
\n" ); document.write( "proof: i/we verify the statement for n=1, i.e (1)_____ -3^1, divisible 5
\n" ); document.write( "II. assume that the statement is there for some integer n=k
\n" ); document.write( "(2)_____ is divisible by 5
\n" ); document.write( "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.
\n" ); document.write( "consider: 8k+1-3k+1=8k*8-3k*3
\n" ); document.write( " =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)_____.\r
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Algebra.Com's Answer #798827 by math_helper(2461)\"\" \"About 
You can put this solution on YOUR website!

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document.write( "They are illustrating proof by induction
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document.write( "(1)  8-3  is divisible by 5  (should not have ^1)
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document.write( "(2)  8k-3k  is divisible by 5  (this is the hypothesis)
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document.write( "(3) Let n=k+1; then...
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document.write( "8(k+1) - 3(k+1) 
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document.write( "= 8k+8 - (3k+3) 
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document.write( "= 8k+8 -3k-3
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document.write( "= (8k-3k)+5 \r
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document.write( "By hypothesis, 8k-3k is divisible by 5  (that was the n=k case)\r
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document.write( "so  8k-3k+5 is also divisible by 5 (if  F is divisible by 5, so is F+5)\r
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document.write( "-----
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document.write( "This part:
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document.write( "\"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.
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document.write( "consider: 8k+1-3k+1=8k*8-3k*3\"\r
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document.write( "should be:
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document.write( "\"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.
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document.write( "Consider:  8(k+1)-3(k+1) = 8k+8-3k-3\"  \r
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document.write( "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.  \r
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document.write( "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.\r
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document.write( "Prove  S = 1+2+3+...+n = (1/2)(n+1)(n)  by induction \r
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document.write( "n=1:  S=1   and  (1/2)(1+1)(1) = 2/2 = 1   (holds for n=1, this is the base case)
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document.write( "Assume true for n=k  (*)   (the hypothesis)\r
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document.write( "Let n=k+1:   S = [ 1+2+...+k ] + (k+1) = [ (1/2)(k+1)(k) ] + (k+1)
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document.write( "where we applied the hypothesis (*) for bracketed [ ] items.\r
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document.write( "Factor out k+1:  S = (k+1) ((1/2)(k) + 1)\r
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document.write( "         since  (1/2)k + 1 =  k/2 + 2/2 =  (k+2)/2  we end up with\r
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document.write( "     S = (k+1)(k+2)/2 =  (1/2)(k+1)(k+2)  
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document.write( "( if you write this in terms of n you have S = (1/2)(n)(n+1) )  \r
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document.write( "thus we have shown the hypothesis holds true for n=k+1 and the proof is complete.\r
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document.write( "...and if you wish, by direct proof:\r
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document.write( "   write the su              S = 1 +  2 +    3 +...+ (n-1) + n
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document.write( "   also write it             S = n +(n-1)+(n-2) +...  + 2  + 1
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document.write( "                            -----------------------------------
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document.write( "   add both eqns            2S = (n+1)+(n+1)+(n+1)+...+(n+1)+(n+1)\r
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document.write( "The RHS has n terms of n+1, that's just n*(n+1):
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document.write( "                            2S = n*(n+1)
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document.write( "              and finally    S = (1/2)*n*(n+1)       
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