Take two consecutive terms kth and (k+1)th: k(k+1)/2 + (k-1)k/2=(k^2+k+k^2-k)/2=k^2. So, taken in pairs, the sum=sum of squares of natural numbers. [1+(1+2)]=2^2; [(1+2)+(1+2+3)]=3^2, etc.
Let's work out the first few terms of the series: 1, 1+3=4, 1+3+6=10, 1+3+6+10=20, 1+3+6+10+15=35,...
Take the series: 1, 4, 10, 20, 35, 56,... Now work out the difference between terms: 3, 6, 10, 15, 21, ... Then the next difference between these: 3, 4, 5, 6, ... And finally the next differences: 1, 1, 1,,...
The constant 1 appears after three differentials, so we can write S=An^3+Bn^2+Cn+D. D=0 because S=0 when n=0. Now we have a system of equations created by plugging in n=1, 2, 3:
(1) A+B+C=1; (2) 8A+4B+2C=4; (3) 27A+9B+3C=10.
(2)-2(1)=6A+2B=2, so (4) 3A+B=1 and B=1-3A
(3)-3(1)=24A+6B=7. Substitute for B from (4): 24A+6-18A=7; 6A+6=7; 6A=1, so A=1/6 and B=1-1/2=1/2. Therefore C=1-(1/2+1/6)=1/3.
S=n^3/6+n^2/2+n/3=n(n^2+3n+2)/6=n(n+1)(n+2)/6. The general term T=the sum of the natural numbers=n(n+1)/2.
CHECK: put n=4: 4*5*6/6=20, the 4th term; put n=5: 5*6*7/6=35, the 5th term.