已知数列｛An｝的前n项和Sn＝1-nAn（n∈正整数）（1）计算A1,A2,A3,A4；（2）猜想An的表达式,并用数学归纳法证明
2019-04-14
已知数列｛An｝的前n项和Sn＝1-nAn（n∈正整数）
（1）计算A1,A2,A3,A4；（2）猜想An的表达式,并用数学归纳法证明
优质解答
n=1时 S1 = 1 - A1 = A1A1 = 1/2n≥2时Sn = 1 - nAnSn-1 = 1 - (n-1)An-1 两式相减An = (n-1)An-1 - nAnAn = (n-1)/(n+1) * An-1 (1)A2 = 1/3 * 1/2 = 1/6 = 1/2*3A3 = 2/4 * 1/6 = 1/12 = 1/3*4A4 = 3/5 * 1/12 = 1/20 = 1/4*5A1 = 1/2 = 1/1*2猜想 An = 1/[n(n+1)]则An+1 = 1/[(n+1)(n+2)]An+1/An = n/n+2与(1)式得 An+1 = n/(n+2) An相符所以.略去
n=1时 S1 = 1 - A1 = A1A1 = 1/2n≥2时Sn = 1 - nAnSn-1 = 1 - (n-1)An-1 两式相减An = (n-1)An-1 - nAnAn = (n-1)/(n+1) * An-1 (1)A2 = 1/3 * 1/2 = 1/6 = 1/2*3A3 = 2/4 * 1/6 = 1/12 = 1/3*4A4 = 3/5 * 1/12 = 1/20 = 1/4*5A1 = 1/2 = 1/1*2猜想 An = 1/[n(n+1)]则An+1 = 1/[(n+1)(n+2)]An+1/An = n/n+2与(1)式得 An+1 = n/(n+2) An相符所以.略去