341Is everyone on a holiday or what? That's the only explanation for this one to go unanswered.
Product of any 4 consecutive integers is always 1 less than a perfect square.
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5 Answers
341
62lol.. i guess ;)
atleast the xi class guys should be trying this one ;)
49n(n+1)(n+2)(n+3)
=(n2+3n)(n2+3n+2)
=n4+6n3+9n2+2n2+6n
=n4+6n3+11n2+6n+1-1
=(n2)2 + (3n)2 + (1)2 + 2.(n2).(1) + 2.(3n).(1) + 2.(n2).(3n) - 1
=(n2 + 3n +1)2 - 1
=k2 - 1
thus, n(n+1)(n+2)(n+3) is 1 less than a perfect square!! :D
62n(n+1)(n+2)(n+3)
=(n2+3n)(n2+3n+2)
you made it worse from here
t(t+2) = t2+2t+1-1 = (t+1)2-1
49hehe yea....
didi it in that way...
and just b4 completing just forgot why i had grouped it :P
stupid me! :P