if i have to evaluate :
S=1.2 +2.3+3.4+..................+(n-1).n +n.1
then i have a ready made formula(with proof also)
for evaluating 1.2 +2.3+3.4+..................+(n-1).n as
(n+1)! /(r+1)(n-r)!
putting r=2
we get (n+1)n(n-1) / 3.............
so S=\frac{(n+1)n(n-1)}{3} +n
but using....
(1+2+3+............+n)2 =(12+22+32+.........n2)+2S
evaluating S trough this equation we get a different result.....why is it so ????
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3 Answers
106because ... using the second method involves all pairs of two nos.
there S has nC2 nos (selection of any two out of n)
but the question asks the sum of n such quantities (not all of those pairs)
Further, in the ques, there is a term n(n+1) which is not present in any of the pairs in the second method ... so this method is not applicable.
106again same thing all the terms in the second method are exceeding the ones required
consider 2.4 , which exists in the second method but is not there in the question
