Find the sum to n terms of the series 11⋅2⋅3⋅4+12⋅3⋅4⋅5+13⋅4⋅5⋅6…
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)=1312∗3=118
@moron Yes, you are right. For the first n terms:
∑nk=11k(k+1)(k+2)(k+3)=13(11∗2∗3−1(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[∑n−1k=01(k+1)(k+2)(k+3)−∑nk=11(k+1)(k+2)(k+3)]
=13(11∗2∗3−1(n+1)(n+2)(n+3))
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