Telescoping Series 3 – Identifying The Pattern

380 will have two parts, 19 X20, therefore c = 19 and d =20 Ans

K.K.GARG,India

\cfrac { 1 }{ n(n+1) } =\frac { 1 }{ 380\ } so c=19 and d=20

The problem solve by this equation. {1/(n*(n+1))=(1/n)-(1/n+1)} and the answer is c=19 and d=20

Let c=n d=n+1 Based on the pattern, 1/380 = 1/cd = 1/n(n+1) = 1/n - 1/(n+1) So, we want to solve for n. 1/n(n+1)=1/380 n(n+1) = 380 n^2 + n = 380 n^2+n-380=0 We can plug the quadratic into the quadratic formula, and since we only care about positive solutions we end up with (-1+39)/2 = 38/2 = 19 = n. n=19 n+1=20 By substitution, c=19 d=20

It is simple with the theory. And the theory is n/n(n+1) = (1/n) - (1/n+1) and then the answer will be c = 19 and d = 20.

380 will have two parts, 19 *20, so c = 19 and d =20

the answers are c=19 and d=20 because 1/19-1/20 is 1/380. This is happening because in this series the difference of d and c is only one if it is more the fraction will not simplify fully and we require a one at the top

(1/380)=(1/c)-(1/d)=(d-c)/cd
cd=380 for all answers

The only values that gives d-c=1 is d=20 and c=19.

we know eq. (1/n(n-1))=(1/(n-1))-(1/n),so that if c=19 & d=20 than (1/c-1/d)=1/380

1/19-1/20= (20-19)/19*20= 1/380, so, c=19 and d=20

380 is the product of two numbers, with respect to the pattern given in previous problem, i.e. 19 X 20, therefore c = 19 and d =20