设数列\(\{a_n\}\)的前\(n\)项和\(S_n\)满足\(S_{n+1}=a_2S_n+a_1,\)其中\(a_2\ne 0\)且\(a_2>-1\)
求证:\(S_n\le \dfrac{n}{2}(1+a_n)\) (重庆2012压轴题)

提示:记\(a_2=x\in(-1,0)\cup (0,\infty),a_n=x^{n-1}\),
只需证:\(x^k+x^{n-1-k}\le 1+x^{n-1}\)由因式分解:\((x^k-1)(x^{n-k-1}-1)>0,(k=0,1,\cdots,n-1)\)累加得证.

评:作为一道高考压轴题,这题用倒序相加，和因式分解就能轻松的的解决.