(1)设S=1•C 2 0+2•C 2 1+3•C 2 2又S=3•C 2 2+2•C 2 1+1•C 2 0
相加2S=4(C 2 0+C 2 1+C 2 2)=16,S=8
设S=1•C 4 0+2•C 4 1+3•C 4 2+4•C 4 3+5•C 4 4
又S=5•C 4 4+4•C 4 3+3•C 4 2+2•C 4 1+1•C 4 0
相加2S=6(C 3 0+C 4 1+C 4 2+C 4 3+C 4 4),∴S=3•2 4=48
(2)1•C n 0+2•C n 1+3•C n 2+…+(n+1)C n n=(n+2)•2 n-1
设S=1•C n 0+2•C n 1+3•C n 2+…+(n+1)C n n
又S=(n+1)C n n+nC n n-1+…+1•C n 0
相加2S=(n+2)(C n 0+C n 1+…+C n n)∴ S=
n+2
n • 2 n =(n+2)• 2 n-1
(3)当q=1时S n=na 1S 1C n 0+S 2C n 1+…+S n+1C n n
=a 1C n 0+2a 1C n 1+…+(n+1)a 1C n n
=a 1(1•C n 0+2•C n 1+…+(n+1)C n n)
=a 1•(n+2)•2 n-1
当q≠1时 S n =
a 1 (1- q n )
1-q =
a 1
1-q -
a 1
1-q q n
S 1C n 0+S 2C n 1+S 3C n 2+…+S n+1C n n= (
a 1
1-q -
a 1
1-q q)
C 0n +(
a 1
1-q -
a 1
1-q q 2 )
C 1n +…+(
a 1
1-q -
a 1
1-q q n+1 )
C nn
=
a 1
1-q (
C 0n +
C 1n +…+
C nn )-
a 1
1-q (q
C 0n + q 2
C 1n +…+ q n+1
C nn )
=
a 1
1-q • 2 n -
a 1
1-q •q(
C 0n • q 0 +
C 1n • q 1 +…+
C nn q n )
=
a 1
1-q • 2 n -
a 1
1-q •q(1+q ) n =
a 1 • 2 n
1-q -
a 1 q (1+q) n
1-q
综上,q=1时S 1C n 0+…+S n+1C n n=a 1(n+2)•2 n-1q≠1时 S 1
C 0n +…+ S n+1
C nn =
a 1 • 2 n
1-q -
a 1 q (1+q) n
1-q