Гост написа:Здравейте бихте ли ми помогнали
Намерете Sn на геометричната прогресия, ако
а)а1=1, а10 =512, n=5
б)a3=-8, a6= - 64, n=5
в)a1=3, a5=16/27, n=5
$$ \overset{..}{\underset{..}{-}} a_{1}, a_{2}, a_{3}, ... , a_{k-1}, a_{k}, a_{k+1}, ... , a_{n-1}, a_{n} \hspace{2em} 1<k<n , k \in \mathbb{N}, n \in \mathbb{N}, a_{1}\ne 0$$ $$\boxed{a_{n}=a_{1}\cdot{q^{n-1}}} \Leftrightarrow \boxed{a_{1}=\frac{a_{n}}{q^{n-1}}} \Leftrightarrow \boxed{q^{n-1}=\frac{a_{n}}{a_{1}}}, \hspace{2em} q \in \mathbb{R}, q \ne 0, n \in \mathbb{N}$$ $$ \boxed{a_{k}^{2}=a_{k-1}\cdot{a_{k+1}}}, \hspace{2em} k>1, k \in \mathbb{N} $$ $$ \boxed{a_{1}\cdot{a_{n}}=a_{1+k}\cdot{a_{n-k}}}, \hspace{2em} 1<k<n , k \in \mathbb{N}, n \in \mathbb{N}$$ $$ \boxed{S_{n}=\frac{a_{n}\cdot{q}-a_{1}}{q-1}}, \hspace{2em} q \in \mathbb{R}, q \ne 0, n \in \mathbb{N}$$ $$ \boxed{S_{n}=a_{1}\cdot{\frac{1-q^{n}}{1-q}}}, \hspace{2em} q \in \mathbb{R}, q \ne 0, n \in \mathbb{N} $$ $$ \boxed{q=\sqrt[k-1]{\frac{a_{k}}{a_{1}}}}, \hspace{2em} q \in \mathbb{R}, q \ne 0, k>1, k \in \mathbb{N} $$ $$ \boxed{q^{n-k}=\frac{a_{n}}{a_{k}}}, \hspace{2em} q \in \mathbb{R}, q \ne 0, k>1, k \in \mathbb{N}$$
(а)[tex]a_{1}=1, a_{k}=512, k=10, n=5, q=?, S_{n}=?[/tex]
[tex]q=\sqrt[k-1]{\frac{a_{k}}{a_{1}}}, \hspace{2em} S_{n}=a_{1}\cdot{\frac{1-q^{n}}{1-q}}[/tex]
(б)[tex]a_{3}=-8, a_{6}=-64, n=5, q=?, S_{n}=?[/tex]
[tex]q^{3}=\frac{a_{6}}{a_{3}}, \hspace{2em} a_{1}=\frac{a_{3}}{q^{2}}, \hspace{2em} S_{n}=a_{1}\cdot{\frac{1-q^{n}}{1-q}}[/tex]
(в)[tex]a_{1}=3, a_{k}=\frac{16}{27}, k=5, n=5, q=?, S_{n}=?[/tex]
[tex]q=\sqrt[k-1]{\frac{a_{k}}{a_{1}} }\\ n=k \Rightarrow a_{n}=a_{k} \Rightarrow \hspace{2em} S_{n}=\frac{a_{n}\cdot{q}-a_{1}}{q-1}[/tex]
[tex]q=\pm\frac{2}{3}[/tex]
[tex]\color{lightseagreen}\text{''Който никога не е правил грешка, никога не е опитвал нещо ново.''} \\
\hspace{21em}\text{(Алберт Айнщайн)}[/tex]