
#1
Dec610, 05:24 PM

P: 434

I need to prove by mathematical induction that 6 divides (k^2+5)k for all k>=1. It works for 1 but I am having trouble obtaining the k+1'st term from the k'th term. Can anyone help? Thanks!!




#2
Dec610, 05:35 PM

Sci Advisor
HW Helper
Thanks
P: 26,167

Hi Xyius!
(try using the X^{2} icon just above the Reply box ) show us what you've tried, and where you're stuck, and then we'll know how to help! 



#3
Dec610, 08:54 PM

P: 153

It might be helpful if you restate as k^3 + 5k. Immediately it will be apparent that 2 will divide all values. If 3 also divides all values, then you've got your divisibility by 6.




#4
Dec1810, 10:16 AM

P: 81

Show 6 divides (n^2+5)n
if k=0 mod 6, k^2+5 = 5 mod 6, (k^2+5)k = 0 mod 6
if k=1 mod 6, k^2+5 = 0 mod 6, (k^2+5)k = 0 mod 6 if k=2 mod 6, k^2+5 = 3 mod 6, (k^2+5)k = 0 mod 6 if k=3 mod 6, k^2+5 = 2 mod 6, (k^2+5)k = 0 mod 6 if k=4 mod 6, k^2+5 = 3 mod 6, (k^2+5)k = 0 mod 6 if k=5 mod 6, k^2+5 = 0 mod 6, (k^2+5)k = 0 mod 6 



#5
Jan1411, 02:33 PM

P: 26

(k^2+5)k=k^3+5k
((k+1)^2+5)(k+1)=(k^2+2k+6)(k+1)=k^3+3k^2+8k+6=(k^2+5)k + 3(k^2+k)+6. 3(k^2+k) is a multiple of 6 since k^2+k is even. 



#6
Jan1511, 10:06 AM

Sci Advisor
HW Helper
P: 11,863

You can prove it without mathematical induction, too
[tex] k\left(k^2 +5\right) =k\left[(k+1)(k+2)3k+3\right]= k(k+1)(k+2)3k(k1) [/tex] which is obviously divisible by 6. 



#7
Jan1611, 09:11 AM

P: 399

[tex] f(x)=0mod(n) [/tex] so you try the ansatz [tex] x=imod(n) [/tex] ehere 'i' runs over i=0,1,2,3,4,5,,....,n1 



#8
Jan1711, 02:35 PM

P: 891

[tex]K(K^{2} + 5) = = K(K^{2} 1) + 6[/tex] = [tex](K1)*K*(K+1) + 6[/tex] from which you could deduce that if it is true for k = n then it is obviously true for k = n + 1. or otherwise show as Cubzar did that F(K+1)F(K) [ i.e. 3K^2 + 3K + 6] is always divisible by 6 regardless of the value of K mod 6. [tex](K +1)^{3) +5(K+1) == K^3 + 3K^2 + 8K + 6 


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