Telescoping Series 2 – Future Terms

The problem solve by this equation. {1/(n*(n+1))=(1/n)-(1/n+1)}

it follows a logic so a=11 and b=12

Actually the series is a telescoping series which is a series where each term can be represented in the following way , $u_{k} = t_{k} - t_{k+1} ........(1)$ Where $t_{k}$ is some series.

For this series we can see that $t_{k} = \frac{1}{n}$ so $u_{k} = \frac{1}{n\times(n+1)}$

so from (1) it can be seen that $\frac{1}{n\times(n+1)} = \frac{1}{n} - \frac{1}{n+1}$

The theory is 1/n(n+1))=(1/n)-(1/n+1) and the the answer will be a = 11 and b = 12

when 1/2-1/3 then a=2, b=3 this way 1/11-1/12 gets a=11 b=12

By observing the patter it is like a series with nth term 1/n-1/n+1

its very easy.......i simply looked at the denominator's pattern

I actually just followed the patterns on the denominators.

by solving 1/11-1/12 we get result as 1/(11*12) so a=11& b=12

we know (1/n)-(1/(n+1)) = 1/(n*(n+1)) so in this case n = 11. so a=11 and b=12

same solution because the result has to be positive so the bigger fraction first

Follow the pattern. 1/n - 1/(n+1) = 1/n*(n+1)

The problem can be solved by this equation. {1/(n*(n+1))=(1/n)-(1/n+1)}. Though this is not required because the theory and examples are given. But if they were not there then this equation is the best method.

this problem should have been there without all that theory and examples thers nothing to do just write 1 as 12-11 and cancel

1/11- 1/12 ,shows a=11 and b=12 Ans

similar to given details, 1/11- 1/12 ,shows a=11 and b=12 Ans

K.K.GARG,India