题目

已知数列{an}的前n项和为Sn，且．(Ⅰ) 求证：数列{an＋1}为等比数列； (Ⅱ) 令bn=，求数列{bn}的的前n项和Tn. 答案：（I）由，可得S1＝2a1－1，即a1＝1，······················································· 1分 又因为， 相减得 即································································ 2分 所以， 故{an＋1}是以a1＋1＝2为首项，以2为公比的等比数列11．先化简$\frac{{{x^2}+2x+1}}{2x-6}÷（x-\frac{1-3x}{x-3}）$，并回答：原代数式的值可能等于1吗，为什么？