求证：1\a(a+d)+1\(a+d)(a+2d)+.+1\[a+(n-2)d][a+(n-1)d]=(n-1)\a[a+(n-1)d]

Question

求证：1\a(a+d)+1\(a+d)(a+2d)+.+1\[a+(n-2)d][a+(n-1)d]=(n-1)\a[a+(n-1)d]

Answer 1

/是分号,*是乘号
1/a(a+d)
=1/d*(1/a-1/(a+d))
1/(a+d)(a+2d)
=1/d*(1/(a+d)-1/(a+2d))
.
1/a(a+d)+1/(a+d)(a+2d)+.+1/[a+(n-2)d][a+(n-1)d]
=1/d*(1/a-1/(a+d)1+1/(a+d)-1/(a+2d)+.+1/[a+(n-2)d]-1/[a+(n-1)d])
=1/d*(1/a-1/[a+(n-1)d])
=1/d*{(a+(n-1)d-a)/[a(a+(n-1)d)]}.内部通分,分子是a+(n-1)d-a=(n-1)d
=1/d*(n-1)d/[a(a+(n-1)d)]
=(n-1)/[a(a+(n-1)d)]
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