a question from progression

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 ????

3 Answers

106
Asish Mahapatra ·

because ... 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.

106
Asish Mahapatra ·

again 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

1
xYz ·

@ashish.ur rit.i realise my mistake........thanx :)

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