Với n = 1 thì \(x^1\ge2.x^0=0\)

Giả sử đẳng thức đúng với n = k nghĩa là : \(x^k\ge\left(k+1\right).x^{k-1}\).

Ta phải chứng minh :

\(x^n\ge\left(n+1\right).x^{n-1}\)đúng với n = k + 1. Ta phải chứng minh \(x^{k+1}\ge\left[\left(k+1\right)+1\right].x^{\left(k-1\right)+1}=\left(k+2\right).x^k\)

\(=\left(x^k.k+2x^k+1\right)-1=\left(x^k+1\right)^2-1\le x^{k+1}\)

Vậy đẳng thức luôn đúng với mọi \(n\inℕ^∗\)

\(P=\frac{sina+cosa}{sina-cosa}=\frac{\frac{sina}{sina}+\frac{cosa}{sina}}{\frac{sina}{sina}-\frac{cosa}{sina}}=\frac{1+cota}{1-cota}=\frac{1+2}{1-2}=-3\)

Với \(n=0\Rightarrow0-0+0-0+0-0=0⋮24\left(đúng\right)\)

Với \(n=1\Rightarrow1-3+6-7+5-2=0⋮24\left(đúng\right)\)

G/s \(n=k\Rightarrow\left(k^6-3k^5+6k^4-7k^3+5k^2-2k\right)⋮24\)

\(\Rightarrow k\left(k^5-3k^4+6k^3-7k^2+5k-2\right)⋮24\\ \Rightarrow k\left(k+1\right)\left(k^2+k+1\right)\left(k^2-k+2\right)⋮24\)

Với \(n=k+1\), ta cần cm \(\left[\left(k+1\right)^6-3\left(k+1\right)^5+6\left(k+1\right)^4-7\left(k+1\right)^3+5\left(k+1\right)^2-2\left(k+1\right)\right]⋮24\)

Ta có \(\left(k+1\right)^6-3\left(k+1\right)^5+6\left(k+1\right)^4-7\left(k+1\right)^3+5\left(k+1\right)^2-2\left(k+1\right)\)

\(=\left(k+1\right)\left[\left(k+1\right)^5-3\left(k+1\right)^4+6\left(k+1\right)^3-7\left(k+1\right)+5\left(k+1\right)-2\right]\\ =\left(k+1\right)\left(k+1-1\right)\left[\left(k+1\right)^2-\left(k+1\right)+1\right]\left[\left(k+1\right)^2-\left(k+1\right)+2\right]\\ =k\left(k+1\right)\left(k^2+k+1\right)\left(k^2+k+2\right)\)

Mà theo GT quy nạp ta có \(k\left(k+1\right)\left(k^2+k+1\right)\left(k^2+k+2\right)⋮24\)

Vậy ta được đpcm