Wednesday, June 27, 2018

calculus - How can the following be calculated?



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

(20121)(1)+(20122)(2)+(20123)(3)++(20122011)(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)22011(2011+1)(22011+1)6=1357477286.






This also evaluates the general sum:
1+(1+2)++(1+2++n)

=(n+11)(1)+(n+12)(2)++(n+1n)(n)
=(n+1)(1+2++n)(12+22++n2)
=(n+1)n(n+1)2n(n+1)(2n+1)6=n(n+1)[n+122n+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+131]
=n(n+1)(n+2)6.


No comments:

Post a Comment

analysis - Injection, making bijection

I have injection f:AB and I want to get bijection. Can I just resting codomain to f(A)? I know that every function i...