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．

Question

设数列{a_n}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

．

Answer 1

由数列{a_n}是公差为d，且首项为a₀=d的等差数列
得：a_n=a₀+（n+1-1）d=（n+1）d；
∴Sn+1＝a0

C

0 n

+a1

C

1 n

+…+an

C

n n

，
又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．