Soluzione - Sequenze geometriche

Per calcolare la forma generale della serie, inserisci il primo termine: $$a=43$$ e il rapporto comune: $$r=1,3255813953488371$$ nella formula per le serie geometriche:

$$a_1=43$$

$$a_2=a_1·r^{n-1}=43\cdot 1,{3255813953488371}^{2-1}=43\cdot 1,{3255813953488371}^1=43\cdot 1,3255813953488371=56,99999999999999$$

$$a_3=a_1·r^{n-1}=43\cdot 1,{3255813953488371}^{3-1}=43\cdot 1,{3255813953488371}^2=43\cdot 1,75716603569497=75,55813953488371$$

$$a_4=a_1·r^{n-1}=43\cdot 1,{3255813953488371}^{4-1}=43\cdot 1,{3255813953488371}^3=43\cdot 2,329266605456123=100,15846403461329$$

$$a_5=a_1·r^{n-1}=43\cdot 1,{3255813953488371}^{5-1}=43\cdot 1,{3255813953488371}^4=43\cdot 3,0876324769999766=132,768196510999$$

$$a_6=a_1·r^{n-1}=43\cdot 1,{3255813953488371}^{6-1}=43\cdot 1,{3255813953488371}^5=43\cdot 4,092908167186016=175,99505118899867$$

$$a_7=a_1·r^{n-1}=43\cdot 1,{3255813953488371}^{7-1}=43\cdot 1,{3255813953488371}^6=43\cdot 5,425482919293089=233,29576552960285$$

$$a_8=a_1·r^{n-1}=43\cdot 1,{3255813953488371}^{8-1}=43\cdot 1,{3255813953488371}^7=43\cdot 7,191919218597816=309,2525263997061$$

$$a_9=a_1·r^{n-1}=43\cdot 1,{3255813953488371}^{9-1}=43\cdot 1,{3255813953488371}^8=43\cdot 9,533474313025012=409,9393954600755$$

$$a_{10}=a_1·r^{n-1}=43\cdot 1,{3255813953488371}^{10-1}=43\cdot 1,{3255813953488371}^9=43\cdot 12,637396182381991=543,4080358424256$$