Question

已知数列｛an｝满足a1=1，a2=2，a3=3，a4=4，a5=5，当n≥5时，a(n+1)=a1a2……an-1若数列{bn

已知数列｛an｝满足a1=1，a2=2，a3=3，a4=4，a5=5，当n≥5时，a(n+1)=a1a2……an-1，若数列{bn}（n∈N*）满足bn=a1a2……an-a1^2-a2^2-……-an^211.求b5 2,求证n＞5时bn+1-bn=1

Answer 1

解：由已知，b5=a1a2…a5-a12-a22-…-a52
=1×2×3×4×5-（12+22+32+42+52）
=120-55
=65．
当n≥5时，由an+1=a1a2…an-1，移向得出a1a2…an-1=an+1+1 ①
∵bn=a1a2…an-a12-a22-…-an2，②
∴bn+1=a1a2…anan+1-a12-a22-…-an2-an+12 ③
③-②得bn+1-bn=a1a2…anan+1-a1a2…an-an+12
=a1a2…an（an+1-1）-an+12 （将①式代入）
=（an+1+1）（an+1-1）-an+12=an+12-1-an+12
=-1

Answer 2

b（n+1）-bn=a1a2......ana(n+1) - a1a2...an - a(n+1)^2
a1a2...an = a(n+1) + 1