Soluzione - Sequenze geometriche

Per calcolare la forma generale della serie, inserisci il primo termine: $$a=903$$ e il rapporto comune: $$r=1,1229235880398671$$ nella formula per le serie geometriche:

$$a_1=903$$

$$a_2=a_1·r^{n-1}=903\cdot 1,{1229235880398671}^{2-1}=903\cdot 1,{1229235880398671}^1=903\cdot 1,1229235880398671=1014$$

$$a_3=a_1·r^{n-1}=903\cdot 1,{1229235880398671}^{3-1}=903\cdot 1,{1229235880398671}^2=903\cdot 1,2609573845763293=1138,6445182724253$$

$$a_4=a_1·r^{n-1}=903\cdot 1,{1229235880398671}^{4-1}=903\cdot 1,{1229235880398671}^3=903\cdot 1,4159587906538182=1278,6107879603978$$

$$a_5=a_1·r^{n-1}=903\cdot 1,{1229235880398671}^{5-1}=903\cdot 1,{1229235880398671}^4=903\cdot 1,5900135257175767=1435,7822137229718$$

$$a_6=a_1·r^{n-1}=903\cdot 1,{1229235880398671}^{6-1}=903\cdot 1,{1229235880398671}^5=903\cdot 1,7854636933307009=1612,273715077623$$

$$a_7=a_1·r^{n-1}=903\cdot 1,{1229235880398671}^{7-1}=903\cdot 1,{1229235880398671}^6=903\cdot 2,0049392968298236=1810,4601850373308$$

$$a_8=a_1·r^{n-1}=903\cdot 1,{1229235880398671}^{8-1}=903\cdot 1,{1229235880398671}^7=903\cdot 2,2513936289982737=2033,0084469854412$$

$$a_9=a_1·r^{n-1}=903\cdot 1,{1229235880398671}^{9-1}=903\cdot 1,{1229235880398671}^8=903\cdot 2,528143011964839=2282,9131398042496$$

$$a_{10}=a_1·r^{n-1}=903\cdot 1,{1229235880398671}^{10-1}=903\cdot 1,{1229235880398671}^9=903\cdot 2,8389114220734735=2563,5370141323465$$