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

Question

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

Answer 1

b(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