1　迭加法的背景若数列 {an}为等差数列 ,则有 an+1 - an= d( n∈ N* ,d为常数 ) .于是 ,an- an- 1 =d,an- 1 - an- 2 =d,… ,a3- a2 =d,a2 - a1 =d,将这 n- 1个式子迭加 ,有 an- a1 =( n- 1 ) d,即得等差数列通项公式 an=a1 + ( n- 1 ) d.考虑到这 n- 1个式子中的被减项是 a2 1 The background of the superposition method If the series {an} is an arithmetic sequence, then there is an+1 - an = d ( n ∈ N*, d is a constant). Then, an- an- 1 = d, an- 1 - an - 2 = d, ..., a3- a2 = d, a2 - a1 = d, superpose these n-1 formulas, and have an-a1 = (n-1) d. An=a1 + ( n- 1 ) d. Consider that the subtracted term in this n-1 formula is a2