Question

Vă rog !!! Dau coroana !!!​

Answer 1

Răspuns:

Explicație pas cu pas:

$a_{n+1} = a_{n} \cdot q$

$S = \dfrac{1}{\dfrac{1}{a_{1}} + \dfrac{1}{a_{2}}} + \dfrac{1}{\dfrac{1}{a_{2}} + \dfrac{1}{a_{3}}} + ... + \dfrac{1}{\dfrac{1}{a_{n}} + \dfrac{1}{a_{n+1}}}=$

$= \dfrac{1}{\dfrac{1}{a_{1}} + \dfrac{1}{a_{1} \cdot q}} + \dfrac{1}{\dfrac{1}{a_{2}} + \dfrac{1}{a_{2} \cdot q}} + ... + \dfrac{1}{\dfrac{1}{a_{n}} + \dfrac{1}{a_{n} \cdot q}}$

$= \dfrac{1}{\dfrac{q + 1}{a_{1} \cdot q}} + \dfrac{1}{\dfrac{q + 1}{a_{2} \cdot q}} + ... + \dfrac{1}{\dfrac{q + 1}{a_{n} \cdot q}} = \dfrac{a_{1} \cdot q}{q + 1} + \dfrac{a_{2} \cdot q}{q + 1} + ... + \dfrac{a_{n} \cdot q}{q + 1}$

$= \dfrac{q \cdot (a_{1} + a_{2} + ... + a_{n})}{q + 1} = \dfrac{q \cdot \dfrac{a_{1}(q^{n} - 1 )}{q - 1} }{q + 1} = \dfrac{a_{1} \cdot q \cdot (q^{n} - 1 )}{(q - 1)(q + 1)}$

Answer 2

$\it \dfrac{1}{\dfrac{1}{a_n}+\dfrac{1}{a_{n+1}}}=\dfrac{1}{\dfrac{a_{n+1}+a_n}{a_n\cdot a_{n+1}}}=\dfrac{a_n\cdot a_{n+1}}{a_{n+1}+a_n}=\dfrac{a_n\cdot a_n\cdot q}{a_n\cdot q+a_n}=\dfrac{a_n\cdot a_n\cdot q}{a_n(q+1)}=\dfrac{q}{q+1}a_n$

$\it S=\dfrac{q}{q+1}a_1+\dfrac{q}{q+1}a_2+\dfrac{q}{q+1}a_3+\ ...\ +\dfrac{q}{q+1}a_n=\\ \\ \\ =\dfrac{q}{q+1}(a_1+a+2+a_3+\ ...\ +a_n)=\dfrac{q}{q+1}\cdot a_1\cdot\dfrac{q^n-1}{q-1}=\dfrac{q(q^n-1)}{q^2-1}a_1$