n i k
: 11 lis 2023, o 15:40
Zwinąć wyrażenie \(\displaystyle{ \frac{ \sum_{k=1}^{n^2-1} \sqrt{n+ \sqrt{k} }}{\sum_{k=1}^{n^2-1} \sqrt{n- \sqrt{k}}}. }\)
\(\displaystyle{ \sqrt{n+ \sqrt{k} } + \sqrt{n- \sqrt{k} } =x_{k} /^2}\)
\(\displaystyle{ 2n+2 \sqrt{n^2-k} =x^2_{k}}\)
\(\displaystyle{ x_{k}= \sqrt{2} \sqrt{n+ \sqrt{n^2-k} } }\)
\(\displaystyle{ \sqrt{n- \sqrt{k} } =x_{k}-\sqrt{n+ \sqrt{k} } }\)
\(\displaystyle{ \sqrt{n- \sqrt{k} } =\sqrt{2} \sqrt{n+ \sqrt{n^2-k} }-\sqrt{n+ \sqrt{k} } }\)
\(\displaystyle{ \frac{ \sum_{k=1}^{n^2-1} \sqrt{n+ \sqrt{k} } }{\sum_{k=1}^{n^2-1} \sqrt{n- \sqrt{k} } } = \frac{\sum_{k=1}^{n^2-1} \sqrt{n+ \sqrt{k} } }{ \sqrt{2}\sum_{k=1}^{n^2-1} \sqrt{n+ \sqrt{n^2-k} }-\sum_{k=1}^{n^2-1} \sqrt{n+ \sqrt{k} } } }\)
ale jak łatwo zauważyć:
\(\displaystyle{ \sum_{k=1}^{n^2-1} \sqrt{n+ \sqrt{n^2-k} }=\sum_{k=1}^{n^2-1} \sqrt{n+ \sqrt{k} } =A}\)
Więc cały ten ułamek to:
\(\displaystyle{ \frac{A}{ \sqrt{2}A-A } = \frac{1}{ \sqrt{2} -1} = \sqrt{2} +1}\)
cnd...