Help! cannot solve P4 problem

[tantrik]

Hi friends,

I got in to this problem from Jan 2002 P4 question paper. The problem is that I need to deduce n(n-1)(2n+5) is divisible by 6 for all n>1. How do I deduce this without induction? This question has only 2 marks.

Thanks in advance.

[Alpha]

I wish I could help...

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[elemis]

Let f(n) = n(n-1)(2n+5)

f(1) = 1(1-1)(2+5) = 0  Since zero is divisible by 6 the statement is proven to be divisible for n = 1

Assuming f(k) is divisible by 6 for all k which are positive integers greater than 1.

Therefore, f(k+1) = (k+1)(k)[2(k+1)+5]

f(k+1) - f(k) = (k+1)(k)[2(k+1)+5] - [k(k-1)(2k+5)]

                  = 2k³ +9k² +7k - 2k³ -3k² +5k
              
                  =6k² + 12
 
                  = 6(k² +2)

Hence,         f(k+1) = 6(k² +2) + f(k)  is proven to be divisible when n=k+1

If f(n) is divisible when n=k it is shown to be divisible when n=k+1 where k is any positive integer greater than 1.
          

[elemis]

I wish I could help...

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No need... Ari is here

[Alpha]

No need... Ari is here

Thank you Lion.

[tantrik]

Thanks Ari. Highly appreciated.

[elemis]

Thanks Ari. Highly appreciated.

No worries.