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设数列{an}是公差为d,且首项为a0=d的等差数列,求和:Sn+1=a0C0n+a1C1n+…+anCnn.

题目详情
设数列{an}是公差为d,且首项为a0=d的等差数列,求和:Sn+1=a0
C
0
n
+a1
C
1
n
+…+an
C
n
n
▼优质解答
答案和解析
由数列{an}是公差为d,且首项为a0=d的等差数列
得:an=a0+(n+1-1)d=(n+1)d;
Sn+1=a0
C
0
n
+a1
C
1
n
+…+an
C
n
n

Sn+1=an
C
n
n
+an−1
C
n−1
n
+…+a0
C
0
n

=an
C
0
n
+an−1
C
1
n
+…+a0
C
n
n

2Sn+1=(a0+an)C
 
0
n
+(a1+an−1)
C
1
n
+…+(an+a0)
C
n
n

=(a0+an)(
C
0
n
+
C
1
n
+…+
C
n
n
)=(a0+an)2n
Sn+1=(a0+an)•2n−1.