Question

DAU COROANA

Aflati suma S=1+$\frac{1}{5}$+$\frac{1}{5^{2} }$+...+$\frac{1}{5^{11} }$

Answer 1

$\it \dfrac{..}{..}\ \ b_1+b_2+b_3+\ ...\ +b_n= b_1\cdot\dfrac{q^n-1}{q-1}\\ \\ \\ 1+\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+\ ...\ +\dfrac{1}{5^{11}}=\ \dfrac{\ \dfrac{1}{5^{12}}-1}{\dfrac{1}{5}-1}=\dfrac{5^{12}-1}{5^{12}}\cdot\dfrac{5}{4}=\dfrac{5^{12}-1}{4\cdot5^{11}}$

Answer 2

Răspuns:

aplici formula invatata pe 1+x+x²+...+x^n

Explicație pas cu pas:

(1-(1/5)^12)/(1-1/5) =((5/4)*(5^12-1))/5^12)=(5^13-5)/4*5^12=(5^12-1)/4*5^11=

5/4-1/(4*5^11)