题目内容
数列{an}和{bn}满足an=
(b1+b2+…+bn)(n=1,2,3…),求证{bn}为等差数列的充要条件是{an}为等差数列.
1 | n |
分析:先证必要性:若{bn}为等差数列,设首项b1,公差d,由题意能导出an+1-an=
,{an}为是公差为
的等差数列.再证充分性若{an}为等差数列,设首项a1,公差d,则能导出bn+1-bn=2d,即{bn}是公差为等差数列.
d |
2 |
d |
2 |
解答:证明:必要性若{bn}为等差数列,设首项b1,公差d
则an=
(nb1+
d)=b1+
d
∵an+1-an=
,∴{an}为是公差为
的等差数列
充分性若{an}为等差数列,设首项a1,公差d
则b1+b2+…+bn=n[a1+(n-1)d]=dn2+(a1-d)
nb1+b2+…+bn-1=d(n-1)2+(a1-d)(n-1),(n≥2)
∴bn=2dn+(a1-2d),(n≥2)
当n=1时,b1=a1也适合
∵bn+1-bn=2d,∴{bn}是公差为2d的等差数列
则an=
1 |
n |
n(n-1) |
2 |
n-1 |
2 |
∵an+1-an=
d |
2 |
d |
2 |
充分性若{an}为等差数列,设首项a1,公差d
则b1+b2+…+bn=n[a1+(n-1)d]=dn2+(a1-d)
nb1+b2+…+bn-1=d(n-1)2+(a1-d)(n-1),(n≥2)
∴bn=2dn+(a1-2d),(n≥2)
当n=1时,b1=a1也适合
∵bn+1-bn=2d,∴{bn}是公差为2d的等差数列
点评:本题考查数列的性质和应用,解题时要注意证明充要性的证明步骤.
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