Question

[Vũ Thảo Vy]

Tính  \(S=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\)\(\frac{1}{25\sqrt{24}+24\sqrt{25}}\)  

Answer 1

[zZz Cool Kid_new zZz]

Ta có:\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n^2+n}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}\)

\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\cdot\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)(Nhân liên hợp)

Áp dụng vào bài toán,ta có:

\(S=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+.....+\frac{1}{25\sqrt{24}+24\sqrt{25}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{100}}\)

\(=\frac{9}{10}\)

Answer 2

[Cố Tử Thần]

1/((k+1)√k+k√(k+1))
=((k+1)√k-k√(k+1))/((k+1)^2.k-k^2(k+1)
=((k+1)√k-k√(k+1))/k(k+1)(k+1-k)
=1/√k-1/√(k+1)
1/(2+√2)=1-1/√2
1/(3√2+2√3)=1/√2-1√3
......
1/(100√99+99√100)=1/√99-1/√100
=>1/(2+√2)+1/(3√2+2√3)+......+1/(100√99+99√100)
=1-1/√2+1/√2-1√3+...+1/√99-1/√100
=1-1/√100=1-1/10=9/10