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#1
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Please help me with my homework! ( induction problem)
Show by induction that (3^(4n+2))+ (5^(2n+1))is a multiple of 14.
First person to do it gets a special mention on my location for a week [img]/images/graemlins/laugh.gif[/img] edit: time limit is 4am today to get a special mention on my location. |
#2
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Re: Please help me with my homework! ( induction problem)
Pick random integer, plug it in, divide result by 14? Or does he want a more rigorous proof?
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#3
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Re: Please help me with my homework! ( induction problem)
how do you do that? [img]/images/graemlins/frown.gif[/img]
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#4
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Re: Please help me with my homework! ( induction problem)
Show for n=1:
3^6+5^3=854 = 14*61 == multiple of 14 Assume for n. Show for n+1: 3^(4(n+1)+2) + 5^(2(n+1)+1) = 3^(4n+6) + 5^(2n+3) = 3^6 * 3^4n + 5^3 * 5^2n = 729(3^4n) + 125(5^2n) = 729 * 81n + 125 * 25n = 62174n = 14*4441n QED |
#5
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Re: Please help me with my homework! ( induction problem)
thanks ! [img]/images/graemlins/laugh.gif[/img]
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#6
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Re: Please help me with my homework! ( induction problem)
[ QUOTE ]
Show for n=1: 3^6+5^3=854 = 14*61 == multiple of 14 Assume for n. Show for n+1: 3^(4(n+1)+2) + 5^(2(n+1)+1) = 3^(4n+6) + 5^(2n+3) = 3^6 * 3^4n + 5^3 * 5^2n = 729(3^4n) + 125(5^2n) = 729 * 81n + 125 * 25n [/ QUOTE ] shouldn't this be: = 729 * 81^n + 125 * 25^n = (9^3)(9^2n) + (5^3)(5^2n) it's not obvious to me how I should continue |
#7
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Re: Please help me with my homework! ( induction problem)
[ QUOTE ]
[ QUOTE ] Show for n=1: 3^6+5^3=854 = 14*61 == multiple of 14 Assume for n. Show for n+1: 3^(4(n+1)+2) + 5^(2(n+1)+1) = 3^(4n+6) + 5^(2n+3) = 3^6 * 3^4n + 5^3 * 5^2n = 729(3^4n) + 125(5^2n) = 729 * 81n + 125 * 25n [/ QUOTE ] shouldn't this be: = 729 * 81^n + 125 * 25^n = (9^3)(9^2n) + (5^3)(5^2n) it's not obvious to me how I should continue [/ QUOTE ] I screwed up. It's not 81n it's 81^n, so we're left with less help than I thought we had. There's more mathnastics to navigate through. Sorry about that! |
#8
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Re: Please help me with my homework! ( induction problem)
Try this ...
1st number (n=1) = 729 + 125 = 854 = 14*61 let a_1 = 729 and b_1 = 125 (n+1)th number = 81a_n + 25b_n (see above) where a_n and b_n are the two terms from the nth number (n+1)th number = 56a_n + 25a_n + 25b_n = 56a_n + 25(a_n+b_n) = 4*14*a_n + 25(a_n+b_n) a_n = an integer a_n+b_n = multiple of 14 therfore (n+1)th number is a multiple of 14 |
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