Hei, trenger litt hjelp med å løse denne...
∞ ∑ k=0. 2^(k+3)/e^(x-3)
Uendelig geometrisk rekke
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mrd
Mente selvfølgelig k og ikke x i eksponenten i nevneren
∞ ∑ k=0. 2^(k+3)/e^(k-3)
EDIT: $\sum_{k=0}^\infty\frac{2^{k+3}}{e^{k-3}} $
∞ ∑ k=0. 2^(k+3)/e^(k-3)
EDIT: $\sum_{k=0}^\infty\frac{2^{k+3}}{e^{k-3}} $
Noen hint:
[tex]\eqalign{ & {2^3}{e^3} + {2^4}{e^2} + {2^5}{e^1} + \cdot \cdot \cdot \cr & \cr & k = \frac{2}{e} < 1 \cr & \cr & \mathop {\lim }\limits_{n \to \infty } {S_n} = \mathop {\lim }\limits_{n \to \infty } a \cdot \frac{{{k^n} - 1}}{{k - 1}} \cr}[/tex]
[tex]\eqalign{ & {2^3}{e^3} + {2^4}{e^2} + {2^5}{e^1} + \cdot \cdot \cdot \cr & \cr & k = \frac{2}{e} < 1 \cr & \cr & \mathop {\lim }\limits_{n \to \infty } {S_n} = \mathop {\lim }\limits_{n \to \infty } a \cdot \frac{{{k^n} - 1}}{{k - 1}} \cr}[/tex]
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