Question

Ambele exercitii din poza, multumesc.
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Answer 1

$217. \\ \\ \displaystyle \sum\limits_{k=1}^{n-1}\dfrac{1}{\sqrt{a_{k+1}}+\sqrt{a_k}} =\sum\limits_{k=1}^{n-1}\dfrac{\sqrt{a_{k+1}}-\sqrt{a_k}}{a_{k+1}-a_k} =\sum\limits_{k=1}^{n-1}\dfrac{\sqrt{a_{k+1}}-\sqrt{a_k}}{a_{k}+q-a_k} = \\ \\ = \sum\limits_{k=1}^{n-1}\dfrac{\sqrt{a_{k+1}}-\sqrt{a_k}}{q} = \dfrac{1}{q}\sum\limits_{k=1}^{n-1} \Big(\sqrt{a_{k+1}}-\sqrt{a_k}\Big)=$

$= \dfrac{1}{q}(\sqrt{a_2}+\sqrt{a_3}+...+\sqrt{a_{n}}-\sqrt{a_1}-\sqrt{a_2}-...-\sqrt{a_{n-1}}) = \\ \\ = \dfrac{\sqrt{a_{n}}-\sqrt{a_{1}}}{q} = \dfrac{a_{n}-a_{1}}{q(\sqrt{a_{n}}+\sqrt{a_{1}})} = \dfrac{a_1+(n-1)q-a_1}{q(\sqrt{a_{n}}+\sqrt{a_{1}})} = \\ \\ \\= \boxed{\dfrac{n-1}{\sqrt{a_{n}}+\sqrt{a_1}}}$

$218.\\ \\ \displaystyle S = \sum\limits_{k=1}^{n-1}\dfrac{a^p_k}{a_{k+1}^p-a_k^p} = \sum\limits_{k=1}^{n-1}\dfrac{(a_1\cdot q^{k-1})^p}{(a_1\cdot q^{k})^p-(a_1\cdot q^{k-1})^p} = \\ \\ =\sum\limits_{k=1}^{n-1}\dfrac{(a_1\cdot q^{k-1})^p}{(a_1\cdot q^{k-1})^p\cdot (q^p-1)} = \sum\limits_{k=1}^{n-1}\dfrac{1}{q^p - 1} =\\ \\\\ = \boxed{\dfrac{n-1}{q^p - 1}}$