Oblicz sumy szeregów:
a)-3 (4/5)^n
b)
a)
zatem wyjściowa suma to -3*4=-12
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a)![\sum_{n=1}^{\infty} (4/5)^n =\frac{4/5}{1-4/5}=4 \sum_{n=1}^{\infty} (4/5)^n =\frac{4/5}{1-4/5}=4](https://tex.z-dn.net/?f=%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D+%284%2F5%29%5En+%3D%5Cfrac%7B4%2F5%7D%7B1-4%2F5%7D%3D4)
zatem wyjściowa suma to -3*4=-12
b)![\sum_{n=1}^{\infty} (4/5)^{2n-1} =\frac{5}{4}*\sum_{n=1}^{\infty} (16/25)^{n} =\frac{5}{4}*\frac{16/25}{1-16/25} =\frac{5}{4}*\frac{16}{9}=\frac{20}{9} \sum_{n=1}^{\infty} (4/5)^{2n-1} =\frac{5}{4}*\sum_{n=1}^{\infty} (16/25)^{n} =\frac{5}{4}*\frac{16/25}{1-16/25} =\frac{5}{4}*\frac{16}{9}=\frac{20}{9}](https://tex.z-dn.net/?f=%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D+%284%2F5%29%5E%7B2n-1%7D+%3D%5Cfrac%7B5%7D%7B4%7D%2A%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D+%2816%2F25%29%5E%7Bn%7D+%3D%5Cfrac%7B5%7D%7B4%7D%2A%5Cfrac%7B16%2F25%7D%7B1-16%2F25%7D+%3D%5Cfrac%7B5%7D%7B4%7D%2A%5Cfrac%7B16%7D%7B9%7D%3D%5Cfrac%7B20%7D%7B9%7D)
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