Who can sum the series
1+(1+2)+(1+2+3)+.....
....+(1+2+....+n)
from the terms 1 to n

Hi!!! we can try to solve ur prob on this line
1
1+2
1+2+3
1+2+3+........+n
-----------------------------
1.n+2(n-1)+3(n-2)+........+n.1
1 wud repeat n times;2 wud be(n-1)times and so on and finally n 1 time only.......
we can further try to consolidate the result
u try. i'll also try if i m able to do

kalhhhhhhh............mpravo paidi mou vivek.....alla me stenoxwrei to gegonos oti shkwnei xeri mono enas....8elw kai apo allous....8ELW KINHSH..XERIA...

no one because its infini we can not sum it.its lim is infini i m sure of that

hello you can just make a comparaison these two terms.it's true that{1+(1+2)+(1+2+3)+...}>= 1+2+3+.......; (because it's a sum of positive terms) and 1+2+3+4+..........=Σ(k)=+∞; where k=1,2,...,+∞.by this comparaison we can see that this sum may not converge in N === Original Message === Who can sum the series
1+(1+2)+(1+2+3)+.....
....+(1+2+....+n)
from the terms 1 to n

1+(1+2)+(1+2+3)+.....
....+(1+2+....+n) (1/2)*n^2*(n+1)

1+(1+2)+(1+2+3)+.....
....+(1+2+....+n)=(1/2)*n^2*(n+1)

1
1+2
1+2+3
1+2+3+........+n =
(1/2)*(sum(k(k+1), k = 1 .. n))

I just've read this and an asnwer is the obvious last answer by Bazmouz, and I haven't seen another more simplified. More intriguing appears to me this facts:

Calculated at hand the first 20 results for the initial cuestion as f(n), as well as it decomposition in prime factors; also I put the function g(n)=n(n+1)/2 for the later comment:
f(1)=1 1 g(1)=1 1
f(2)=4 2*2 g(2)=3 3
f(3)=10 5*2 g(3)=6 3*2
f(4)=20 5*2*2 g(4)=10 5*2
f(5)=35 7*5 g(5)=15 5*3
f(6)=56 7*2*2*2 g(6)=21 7*3
f(7)=84 7*3*2*2 g(7)=28 7*2*2
f(8)=120 5*3*2*2*2 g(8)=36 3*3*2*2
f(9)=165 11*5*3 g(9)=45 5*3*3
f(10)=220 11*5*2*2 g(10)=55 11*5
f(11)=286 13*11*2 g(11)=66 11*3*2
f(12)=364 13*7*2 g(12)=78 13*3*2
f(13)=455 13*7*5 g(13)=91 13*7
f(14)=560 7*5*2*2*2*2 g(14)=105 7*5*3
f(15)=680 17*2*2*2 g(15)=120 5*3*2*2*2
f(16)=816 17*3*2*2*2*2 g(16)=136 17*2*2*2
f(17)=969 19*17*3 g(17)=153 17*3*3
f(18)=1140 19*5*3*2*2 g(18)=171 19*3*3
f(19)=1330 19*7*5*2 g(19)=190 19*5*2
f(20)=1540 11*7*5*2*2 g(20)=210 7*5*3*2
f(21)=1771 23*11*7 g(21)=231 11*7*3

First can be noticed that odds results are spaced by four, this due to the spacing of evens and odds in (1/2)k(k+1), that are 2 pairs and 2 odds next. Secondly, there is a kind of persistence in the greater prime factor for each number that appear in the descomposition onc it appears, for example the secuence of those prime factors for f(k) is 1,2,5,5,7,7,7,5,11,11,13,13,13,7,17,17,19,19,19,11,23 , that only go down in this first 21 list 3 times (with 5,7,11 between greater ones). These is much more "persisten" than the list for g(k) listed in the right side. I wonder if there can be a sistematic way (as here g(k) generating with a sum f(k)) to form functions that have best "persistence" in the intuitive way described before, and if has to do with study of prime numbers. Any number theorist by here??

As Vivek puts, the series can go like 1.n+2(n-1)+3(n-2)+...+(n-2)3(n-1)2+n.1 taking into account how many times the numbers appear in the sum. for this then we can have the rearrangement n(1+2+...+n)-(2(1)+3(2)+...n(n-1)), that is the same as n^2(n+1)/2-Σ k(k-1) where k changes from 2 till n. Cause the linear term in ths sum we have n^2(n+1)/2 +n(n+1)2-1-Σ k(k-1)=n(n+1)^2/2 -Σ k^2 where in the last equallity the 1 is introduced in the sum such that k nows ranges from 1 to n. And now this n(n+1)^2)/2 -Σ k^2 is nicer than the last post since the only sum is over squares, for example for n=10, the sum of squares gives 385 and (10(11)^2)/2 is 605, so we got 220 again.