How can the following series be calculated?
S=1+(1+2)+(1+2+3)+(1+2+3+4)+⋯+(1+2+3+4+⋯+2011)
Answer
Note that 1 occurs 2011 times; 2 occurs 2010 times; 3 occurs 2009 times, and so on, until 2011 occurs only once. Hence we can rewrite the sum as
(2012−1)(1)+(2012−2)(2)+(2012−3)(3)+⋯+(2012−2011)(2011).
Split and regroup terms:
2012(1+2+3+⋯+2011)−(12+22+32+⋯+20112).
Now using the two formulas for triangular numbers and square pyramidal numbers, compute
20122011(2011+1)2−2011(2011+1)(2⋅2011+1)6=1357477286.
This also evaluates the general sum:
1+(1+2)+⋯+(1+2+⋯+n)
=(n+1−1)(1)+(n+1−2)(2)+⋯+(n+1−n)(n)
=(n+1)(1+2+⋯+n)−(12+22+⋯+n2)
=(n+1)n(n+1)2−n(n+1)(2n+1)6=n(n+1)[n+12−2n+16]
=n(n+1)(n+2)6.
One could also use the triangle number formula on each term for a more direct route:
1(1+1)2+2(2+1)2+3(3+1)2+⋯+n(n+1)2
=12[(1+22+⋯+n2)+(1+2+⋯+n)]
=12[n(n+1)(2n+1)6+n(n+1)2]=n(n+1)4[2n+13−1]
=n(n+1)(n+2)6.
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