Chứng minh rằng với mọi n $\in$ ℕ^* ta có:

a) $\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\dots +\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n+1}-1.$

b) $\frac{2^3-1}{2^3+1}\cdot \frac{3^3-1}{3^3+1}\cdot \frac{4^3-1}{4^3+1}\cdots \frac{n^3-1}{n^3+1}=\frac{2\left(n^2+n+1\right)}{3n(n+1)}.$

Vậy mệnh đề cũng đúng với n = k + 1. Do đó theo nguyên lí quy nạp toán học, mệnh đề đã cho đúng với mọi n $\in$ ℕ^*.

b)

+) Khi n = 2, ta có:

$\frac{2^3-1}{2^3+1}=\frac{7}{9}=\frac{2\left(2^2+2+1\right)}{3.2(2+1)}.$

Vậy mệnh đề đúng với n = 2.

+) Với k là một số nguyên dương tuỳ ý mà mệnh đề đúng, ta phải chứng minh mệnh đề cũng đúng với k + 1, tức là:

$\frac{2^3-1}{2^3+1}\cdot \frac{3^3-1}{3^3+1}\cdot \frac{4^3-1}{4^3+1}\cdots \frac{{\left(k+1\right)}^3-1}{{\left(k+1\right)}^3+1}=\frac{2\left[{\left(k+1\right)}^2+\left(k+1\right)+1\right]}{3\left(k+1\right)\left[\left(k+1\right)+1\right]}.$ 

Thật vậy, theo giả thiết quy nạp ta có:

$\frac{2^3-1}{2^3+1}\cdot \frac{3^3-1}{3^3+1}\cdot \frac{4^3-1}{4^3+1}\cdots \frac{k^3-1}{k^3+1}=\frac{2\left(k^2+k+1\right)}{3k(k+1)}.$

Khi đó:

$\frac{2^3-1}{2^3+1}\cdot \frac{3^3-1}{3^3+1}\cdot \frac{4^3-1}{4^3+1}\cdots \frac{{\left(k+1\right)}^3-1}{{\left(k+1\right)}^3+1}$

$=\frac{2^3-1}{2^3+1}\cdot \frac{3^3-1}{3^3+1}\cdot \frac{4^3-1}{4^3+1}\cdots \frac{k^3-1}{k^3+1}.\frac{{\left(k+1\right)}^3-1}{{\left(k+1\right)}^3+1}$

$=\left(\frac{2^3-1}{2^3+1}\cdot \frac{3^3-1}{3^3+1}\cdot \frac{4^3-1}{4^3+1}\cdots \frac{k^3-1}{k^3+1}\right).\frac{{\left(k+1\right)}^3-1}{{\left(k+1\right)}^3+1}$

$=\frac{2\left(k^2+k+1\right)}{3k(k+1)}.\frac{{\left(k+1\right)}^3-1}{{\left(k+1\right)}^3+1}$

$=\frac{2\left(k^2+k+1\right)}{3k(k+1)}.\frac{\left[\left(k+1\right)-1\right]\left[{\left(k+1\right)}^2+\left(k+1\right)+1\right]}{\left[\left(k+1\right)+1\right]\left[{\left(k+1\right)}^2-\left(k+1\right)+1\right]}$

$=\frac{2\left(k^2+k+1\right)}{3k(k+1)}.\frac{k\left[{\left(k+1\right)}^2+\left(k+1\right)+1\right]}{\left[\left(k+1\right)+1\right]\left[\left(k^2+2k+1\right)-\left(k+1\right)+1\right]}$

$=\frac{2\left(k^2+k+1\right)}{3k(k+1)}.\frac{k\left[{\left(k+1\right)}^2+\left(k+1\right)+1\right]}{\left[\left(k+1\right)+1\right]\left(k^2+k+1\right)}$

$=\frac{2}{3\left(k+1\right)}.\frac{\left[{\left(k+1\right)}^2+\left(k+1\right)+1\right]}{\left[\left(k+1\right)+1\right]}$

$=\frac{2\left[{\left(k+1\right)}^2+\left(k+1\right)+1\right]}{3\left(k+1\right)\left[\left(k+1\right)+1\right]}.$