Telescoping Sequence?

1/ (1 x 2), 1/ (2 x 3), 1/ (3 x 4), 1/ (4 x 5), .... , 1/ (77 x 78)

Reciprocal is 77 x 78 = 6006

Let the term of the sequence be $a_n=\dfrac{1}{b_n}$ . We note that $b_n = n(n+1)$ . Therefore, $b_{77} = 77\times 78 = \boxed{6006}$

factorize the sum by $\frac{1}{2}$ and write the reciprocal of each term to obtain: 1,3,6,10,... which are the triangle numbers. If Tn is the nth triangle number then Tn = $\frac{n(n+1)}{2}$ , but we have already factorized out the 2 , therefore the 77th term = 77 x78 = 6006

flip the sequence and get 2,6,12,20 2 = 1^1+1 6 = 2^2+2 12=3^2 + 3 20=4^2 +4 this is of the form n^2+n.

this is the reciprocal as it is. Therefore, the reciprocal of the 77th term is 77^2+77 = 6006

After googling "telescoping sequence", I found on Wikipedia that the sought out function is f(x)=1/(x(x+1)). So, plugging it into Python...

1
2
3
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from fractions import Fraction as frac
def tele(n):
    return frac(1,n*(n+1))
print tele(77)**-1

... I find that the reciprocal of f(77) is 6006.

By intuition the Tn of the reciprocal of thee series can be stated as n^2 + n. Therefore 77^2 + 77 = 6006

Result is 1/6006, not 6006, probably sequence was 2,6,12,20,first 3 number are number of my date of birth,so it is interesting for me.6006=77*78