Geometric SeriesIn Probability, ∑j=kNaj=ak−aN+11−a\sum\limits_{j=k}^{N} a^j = {{a^k - a^{N+1} \over {1-a}}}j=k∑Naj=1−aak−aN+1 if ∣a∣<1, ∑j=k∞aj=limN→∞∑j=kNaj=ak1−a~\text{if}~ |a| < 1, ~\sum\limits_{j=k}^{\infty} a^j = \lim_{N \to \infty} \sum\limits_{j=k}^{N} a^j = {{a^k \over {1-a}}} if ∣a∣<1, j=k∑∞aj=N→∞limj=k∑Naj=1−aak ∑j=1∞aj=a1−a\sum\limits_{j=1}^{\infty} a^j = {a \over {1-a}}j=1∑∞aj=1−aa ∑j=0∞aj=a1−a\sum\limits_{j=0}^{\infty} a^j = {a \over {1-a}}j=0∑∞aj=1−aa ∑j=1∞jaj=a(1−a)2\sum\limits_{j=1}^{\infty} ja^j= {a \over {(1-a)^2}}j=1∑∞jaj=(1−a)2a ∑j=0∞jaj=a(1−a)2\sum\limits_{j=0}^{\infty} ja^j= {a \over {(1-a)^2}}j=0∑∞jaj=(1−a)2a ∑j=1∞j2aj=a+a2(1−a)2\sum\limits_{j=1}^{\infty} j^2 a^j= {{a + a^2} \over {(1-a)^2}}j=1∑∞j2aj=(1−a)2a+a2 ∑j=0∞j2aj=a+a2(1−a)2\sum\limits_{j=0}^{\infty} j^2 a^j= {{a + a^2} \over {(1-a)^2}}j=0∑∞j2aj=(1−a)2a+a2Backlinks (1)230131Comments (0)
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