IIT JEE , AIEEE Forum - Mathematics , Algebra

jacques_kallis

Find the sum of the n terms of the series

1+ 2+2+3+3+3+4+4+4+4+5+5+5+5+5+6...........

nrki99

consider an rth term then take sigma from 1 to n

Asmita

1+ 2+2+3+3+3+4+4+4+4+5+5+5+5+5+6...........

=1+2(2)+3(3)+..........+n(n)

=1^2+2^2+3^2+....+n^2

=n(n+1)(2n+1)/6

jacques_kallis

thanks a lot

shakirshafi12

asmita check again not valid for intermediate terms

Asmita

What u said is not clear 2 me.

shakirshafi12

ok suppose n=2
according 2 ur formula
sum=2(3)(5)/6=5 =1^2+2^2
thats why i am saying so (i have done this problem and know the answer)

Asmita

If n=2

Then the sum=1+2+2=5

Which is in accordance with my answer.

shakirshafi12

no in the question
1+ 2+2+3+3+3+4+4+4+4+5+5+5+5+5+6...........
when n=2 then sum should be 1+2=3

nrki99

hi shakirshafi,

i think the representation of question is not correct

acc to me if 1+(2+2)+ --------------------------+(n+n+ ----------------n times)

then the answer given by asmita is correct

& if there is

1+2+2----------------------------------- till n terms

then answr should be different

i am not sure please tell if i am incorrect

shakirshafi12

oh no now mr jacques kallis has the wrong answer !

shakirshafi12

nrki99 this is a fairly cmn question and
1+2+2----------------------------------- till n terms
the answer should be different

shakirshafi12

o h yes!
because mr jacques kallis is happy with the solution may be the question he typed is wrongly represented.

meghaguptaa

May I kno what the ans is? Don't u 2 want 2 kno Asmita??