Question

stie cineva de pe aici ?​

Answer 1

$\dfrac{b_1+b_2}{b_2+b_3}+\dfrac{b_2+b_3}{b_3+b_4}+...+\dfrac{b_{n-1}+b_n}{b_n+b_{n+1}} = \\ \\= \displaystyle \sum\limits_{k=1}^{n-1}\dfrac{b_k+b_{k+1}}{b_{k+1}+b_{k+2}} = \sum\limits_{k=1}^{n-1}\dfrac{b_1\cdot q^{k-1}+b_1\cdot q^k}{b_1\cdot q^k+b_1\cdot q^{k+1}} = \\ \\ =\sum\limits_{k=1}^{n-1}\dfrac{b_1\cdot q^k\cdot (q^{-1}+1)}{b_1\cdot q^k\cdot (1+q)} =\sum\limits_{k=1}^{n-1}\dfrac{q^{-1}+1}{q+1} =$

$\displaystyle = \sum\limits_{k=1}^{n-1}\dfrac{q(q^{-1}+1)}{q(q+1)} =\sum\limits_{k=1}^{n-1}\dfrac{1+q}{q(q+1)} = \sum\limits_{k=1}^{n-1}\dfrac{1}{q} = \\ \\ \\ = (n-1)\cdot \dfrac{1}{q} = \boxed{\dfrac{n-1}{q}}$

=> e) corect