Question

Find the value of :


\[\frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \frac{1}{4 \times 5} + \frac{1}{5 \times 6} + ............+ \frac{1}{9 \times 10} \]


Explain your answer....

Answer 1

what is the '..............'?

Answer 2

It means the sum goes to 1/(9*10....

Answer 3

The terms in place of ............................. are:

\[\frac{1}{6 \times 7} + \frac{1}{7 \times 8} + \frac{1}{8 \times 9}\]

Answer 4

We have
\[\frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \frac{1}{4 \times 5} + \frac{1}{5 \times 6} + ............+ \frac{1}{9 \times 10}\]
first term can be written as
\[\frac{1}{2\times 3}=\frac 12 -\frac 13\]
Using this we'll rewrite the termsas
\[\frac{1}{2}-\frac 13+\frac 13-\frac 14+\frac 14-\frac 15+\frac 15-\frac 1 6+\frac 16-\frac 17+\frac 17-\frac 18+\frac 18-\frac 19+\frac19-\frac 1{10}\]

Now we'll get finally
\[\frac 12-\frac {1}{10}\]
I think you can solve now!

Answer 5

Is 7429 is a factor of 20160??

Answer 6

2/5..

Yeah thanks @ash2326 a lot..

Answer 7

Did you understand @waterineyes ?

Answer 8

Ya I did understand..
\[\frac{1}{5 \times 6} = \frac{1}{5} - \frac{1}{6} = \frac{6-5}{30} = \frac{1}{ 5 \times 6}\]

Answer 9

So every term in between cancels out each other and we are left with first and the last term...

Answer 10

Yeah
\[\frac{1}{2}-\cancel{\frac 13}+\cancel{\frac 13}-\frac 14+\frac 14-\frac 15+\frac 15-\frac 1 6+\frac 16-\frac 17+\frac 17-\frac 18+\frac 18-\frac 19+\frac19-\frac 1{10}\]

Answer 11

Yeah I got it ashutosh..
Thanks...

Answer 12

Welcome:D