Question

tolong bantu jawab kak​

Answer 1

Verified answer

[tex]\begin{aligned}\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}}}{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}}&=\frac{\sum^{99}_{k=1}\sqrt{10+\sqrt{k}}}{\sum^{99}_{k=1}\sqrt{10-\sqrt{k}}}\end{aligned}[/tex]

Perhatikan bahwa

[tex]\begin{aligned}\left(\sqrt{10+\sqrt{k}}-\sqrt{10-\sqrt{k}}\right)^2&=(10+\sqrt{k})+(10-\sqrt{k})-2\sqrt{(10+\sqrt{k})(10-\sqrt{k})} \\ &=20-2\sqrt{100-k} \\ &=2\times10-2\sqrt{100-k} \\ &=2(10-\sqrt{100-k}) \\ \left(\sqrt{10+\sqrt{k}}-\sqrt{10-\sqrt{k}}\right)&=\sqrt{2(10-\sqrt{100-k})} \\ &=\sqrt{2}\left(\sqrt{10-\sqrt{100-k}}\right)\end{aligned}[/tex]

Selain itu, tinjau bahwa

[tex]\begin{aligned}\sum^{99}_{k=1}100-k&=\sum_{k=1}^{99}k\end{aligned}[/tex]

maka,

[tex]\begin{aligned}\sum_{k=1}^{99}\left(\sqrt{10+\sqrt{k}}-\sqrt{10-\sqrt{k}}\right)&=\sum_{k=1}^{99}\sqrt{2}\left(\sqrt{10-\sqrt{100-k}}\right) \\ &=\sum_{k=1}^{99}\sqrt{2}\left(\sqrt{10-\sqrt{k}}\right) \\ \sum_{k=1}^{99}\left(\sqrt{10+\sqrt{k}}\right)-\sum_{k=1}^{99}\left(\sqrt{10-\sqrt{k}}\right)&=\sum_{k=1}^{99}\sqrt{2}\left(\sqrt{10-\sqrt{k}}\right)\end{aligned}[/tex]

sehingga, akan didapatkan

[tex]\begin{aligned}\sum_{k=1}^{99}\left(\sqrt{10+\sqrt{k}}\right)&=\sum_{k=1}^{99}\left(\sqrt{10-\sqrt{k}}\right)+\sum_{k=1}^{99}\sqrt{2}\left(\sqrt{10-\sqrt{k}}\right) \\ \sum_{k=1}^{99}\left(\sqrt{10+\sqrt{k}}\right)&=\left(1+\sqrt{2}\right)\left(\sum_{k=1}^{99}\left(\sqrt{10-\sqrt{k}}\right)\right) \\ \frac{\sum_{k=1}^{99}\left(\sqrt{10+\sqrt{k}}\right)}{\sum_{k=1}^{99}\left(\sqrt{10-\sqrt{k}}\right)}&=\left(1+\sqrt{2}\right)\end{aligned}[/tex]

Jadi, hasilnya adalah

[tex]\begin{aligned}1+\sqrt{2}\end{aligned}[/tex]