在二项式定理这节教材中有这样一个性质:C n 0 +C n 1 +C n 2 +C n 3 +…C n n =2 n ,

1个回答

  • (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