已知数列 { }满足 =1,
(1)记 = ,写出 , ,并求数列 的通项公式;
(2)求 的前 20项和
( 1)解:由题意得b 1 =a 2 =a 1 +1=2,b 2 =a 4 =a 3 +1=5
∵b 1 =a 2 =a 1 +1,∴a 2 -a 1 =1.
b 2 =a 4 =a 3 +1=a 2 +3 ∴a 4 -a 2 =3.
同理 a 6 -a 4 =3
……
b n =a 2n -a 2n-2 =3.
叠加可知 a 2n -a 1 =1+3(n-1)
∴a 2n =3n-1
∴b n =3n-1.验证可得b 1 =a 2 =2,符合上式.
( 2)解:∵a 2n =a 2n-1 +1
∴a 2n-1 =a 2n -1=3n-2.
∴设{a n }前20项和为S 20
∴S 20 =(a 1 +a 3 +…+a 19 )+(a 2 +a 4 +…+a 20 )
=145+155=300