Sunday, 27 September 2015

algebra precalculus - Find the sum to n terms of the series frac11cdot2cdot3cdot4+frac12cdot3cdot4cdot5+frac13cdot4cdot5cdot6ldots




Find the sum to n terms of the series 11234+12345+13456



Please suggest an approach for this task.


Answer



1k(k+1)(k+2)(k+3)=13(k+3k(k+1)(k+2)(k+3)kk(k+1)(k+2)(k+3))
=13(1k(k+1)(k+2)1(k+1)(k+2)(k+3))



k=11k(k+1)(k+2)(k+3)=13123=118




@moron Yes, you are right. For the first n terms:



nk=11k(k+1)(k+2)(k+3)=13(11231(n+1)(n+2)(n+3))



[edit] more details



nk=11k(k+1)(k+2)(k+3)=nk=113(1k(k+1)(k+2)1(k+1)(k+2)(k+3))
=13[nk=11k(k+1)(k+2)nk=11(k+1)(k+2)(k+3)]
=13[n1k=01(k+1)(k+2)(k+3)nk=11(k+1)(k+2)(k+3)]
=13(11231(n+1)(n+2)(n+3))


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