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Sum of the first n terms of a sequence Given an = 1 / (4N + 1) (4N-1), find SN
A (n) = 1 / [(4N-1) (4N + 1)] = (1 / 2) [1 / (4N-1) - 1 / (4N + 1)] S (n) = (1 / 2) {[1 / 3 + 1 / 7 +...] +1/(4n-1)]-[1/5+1/9+… +1 / (4N + 1)]} so we can't use the method of removing the term, so this sequence has no simple
Summation of the first n terms of a sequence
Sequence: 1 / (1x3), 1 / (3x5), 1 / (5x7) ,1/[（2n-1)(2n+1)]
Well, how about this type?
Split term elimination method
1/(1X3）=（1/1 -1/3）/2
1/(3X5)=（1/3 - 1/5）/2
1/[（2n-1)(2n+1)]=[ 1/(2n-1) - 1/(2n+1)]/2
and so on
Add it again, and the middle term will be eliminated
Let 1 / (1 + root 2), 1 / (root 2 + root 3) The sum of the first n terms of 1 / (root n + root (n + 1)) is SN
an=1/(√n+√n+1)=√(n+1)-√n
So Sn = a1 + A2 + +an
=√2-√1+√3-√2+…… +√(n+1)-√n
=√(n+1)-1
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