若bn=2n+1,求和:b1b2-b2b3+b3b4-b4b5+...+(-1)^(n-1)*bn*bn+1

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  • (n-2)-bn=2(n-2)+1-2n-1=-4

    当n为偶数时:

    b1b2-b2b3+b3b4-b4b5+...+(-1)^(n-1)*bn*b(n+1)

    =b2(b1-b3)+b4(b3-b5)+...+bn[b(n-1)-b(n+1)]

    =-4(b2+b4+...+bn)

    =-4(5+9+...+2n+1)

    =-2n²-6n

    当n为奇数时:

    b1b2-b2b3+b3b4-b4b5+...+(-1)^(n-1)*bn*b(n+1)

    =b2(b1-b3)+b4(b3-b5)+...+b(n-1)[b(n-2)-bn]+bn*b(n+1)

    =-4[b2+b4+...+b(n-1)]+bn*b(n+1)

    =-4(5+9+...+2n-1)+(2n+1)(2n+3)

    =2n²+4n+3