If ak = 1/k-1/(k+1), then find out the summation of a2 to a100?
(Here k, 2 & 100 are suffixes)
a2=3
a3=2
a4=5/3 and so on i dont see any pattern how to find the summation ,is there any direct formula
There is a pattern...
$\displaystyle \sum_{k=2}^{100}\frac{1}{k}-\frac{1}{k+1}=\frac{1}{2}{\color{red}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\dots+\frac{1}{100}}-\frac{1}{101}$
Everything in red cancels out and you're left with $\displaystyle \sum_{k=2}^{100}\frac{1}{k}-\frac{1}{k+1}=\frac{1}{2}-\frac{1}{101}=\frac{99}{202}$.
The way you posted your question could be interpreted in many ways...
$\displaystyle \frac{1}{k}-\frac{1}{k+1}$ or $\displaystyle \frac{1}{k-\frac{1}{k+1}}$ or $\displaystyle \frac{\frac{1}{k-1}}{k+1}$.
To make it easier for us to answer your future questions, please learn some basic $\displaystyle \text{LaTeX}$. See here for help.
Order of operations.
First divide, then subtract.
Also, for the sanity of all of us, use LaTex, or at least use brackets.
As I now see it, you could mean either one of these two:
1/k-1/(k+1) =
$\displaystyle
\frac{1}{k}-\frac{1}{k+1}
$
OR
(1/(k-1))/(k+1) =
$\displaystyle
\frac{\frac{1}{k-1}}{k+1} = \frac{k+1}{k-1}
$
i cant understand
1/(k-1)/(k+1) lets put k=2 now this yields 1/(2-1)/(2+1)=1/3 which is not equal to (2+1)/(2-1)=3 i still cant understand how both the fractions r equal now if we take 1/k-1/(k+1) = 1/2-1/(3)=(3-2)/6 which is equal t0 1/6 now this both r not equal how u say all fractions are equal ..