In the arithmetic sequence an, a (n + 1) = 2n + 1, then Sn = (1 / A1A2) + (1 / a2a3) （1/a99a100）=

In the arithmetic sequence an, a (n + 1) = 2n + 1, then Sn = (1 / A1A2) + (1 / a2a3) （1/a99a100）=

A (n + 1) = 2 (n + 1) - 1, so an = 2N-1, so Sn = 1 / 1 * 3 + 1 / 3 * 5 + +1/197*199=(1/2)(1-1/3)+(1/2)(1/3-1/5)+…… +(1/2)(1/197-1/199)=(1/2)(1-1/3+1/3-1/5+…… +1/197-1/199)=(1/2)(1-1/199)=99/199

Arithmetic sequence an = 2n + 3, sum: (1 / A1A2) + (1 / a2a3) +. + (1 / Anan + 1)

The original formula is 1 / (5 × 7) + 1 / (7 × 9) + 1 / (9 × 11) +. + 1 / [(2n + 3) (2n + 5)]
=1/2[(1/5-1/7)+(1/7-1/9)+(1/9-1/11)+.+1/(2n+3)-1/(2n+5)]
=1/2[1/5-1/(2n+5)]
=n/(10n+25)

If {an} is an arithmetic sequence, an ≠ 0, find 1 / A1A2 + 1 / a2a3 +. + 1 / a (n-1) an
If {an} is an arithmetic sequence, an ≠ 0, find 1 / A1A2 + 1 / a2a3 +. + 1 / a (n-1) an
Using the elimination method to do

Let A1 = a
Then 1 / a 1A2 + 1 / a 2A3 +. + 1 / a (n-1) an
=1/a(a+d)+1/(a+d)(a+2d)+…… +1/[a+(n-2)d][a+(n-1)d]
={d/a(a+d)+d/(a+d)(a+2d)+…… +d/[a+(n-2)d][a+(n-1)d]}/d
={1/a-1/(a+d)+1/(a+d)-1/(a+2d)+…… +1/[a+(n-2)d]-1/[a+(n-1)d]}/d
={1/a-1/[a+(n-1)d]}/d
=[1/a-1/(a+nd-d)]/d
=(a+nd-d-a)/d(a+nd-d)
=(nd-d)/d(a+nd-d)
=(n-1)/(a+nd-d)