## how to solve this sort of questions

Find the following sum: 1 /(1.2) + 1 /(2.3)+ 1 /(3.4) + . . . . .+ 1 / (99.100)

## Yes I am getting the same

Yes I am getting the same answer This can also be done in a much simple and lengthy method 1/(1.2) = 1/2 1/(1.2) + 1/(2.3) =1/2 + 1/6 = 2/3 1/(1.2) + 1/(2.3) + 1/(3.4) = 2/3 + 1/12 = 9/12 = 3/4 . . . . . .. . . . . . . . . . similiarly for any number n 1/(1.2) + 1/(2.3) + 1/(3.4)+ . . . . +1/[(n-1).n] = (n-1)/n

## soln fa ya problmm

very easy kin offf..... 1/(1.2)+1/(2.3)+........+(1/99.100) sol: jes analyse dis 1/(1.2) can be written as 1/1-1/2 similarly 1/(2.3) can be written as 1/2-1/3 . . . . . . . similarly 1/(99.100) acn be written as 1/99-1/100 so whole eqn becomes (1/1-1/2)+(1/2-1/3)+(1/3-1/4)+..........+(1/99-1/100) 1/2,1/3,1/4..........1/99 gets cancelled out..remainin terms are........ 1/1-1/100 ans is 99/100 is it d ans???

## let me give you the method

let me give you the method of partial fractions which is the same as what ihim did but only more general. 1/(x+a)(x+b)()x+c).... can be written as A/(x+a) + B/(x+b) + ...... using a simple rule of thumb you can find A,B and so on eg. to find A take -a and substitute in the original fraction after removing (x+a) of course.Similarly for B substitute -b and so on. here : 1/n(n+1) = A/n + B/(n+1) to find A subst. 0 in 1/(n+1) . Thus A = 1 to find B subst. -1 in 1/n . Thus B = -1 thus 1/n(n+1) = 1/n - 1/(n+1) giving the general term. now you can go ahead as ihim....!

## not suggestable...

cum on rajorshi...!!!! dat method iz nt suggestable..though its havin da basic mantra behind d question....!!!

## You must have mathematics in

You must have mathematics in your graduation level thatâ€™s why in the first look you can determine what to write in place 1/1.2 For beginnerâ€™s rajorshi's solutions given a good explanation and to start with they shd solve like that only.

## well ! isnt learning new

well ! isnt learning new things good ! as to how one can apply them totally depends on the individual , right :) also the point should be how to solve problems of any complexity once we can find the pattern. Also by learning techniques you can depend a LOT LESS on providence during the exam as to whether the solution "strikes you or not". And look at it this way, if for every question you have to start from scratch, then what is the use of practicing so many problems. I should not be just for the confidence 'cause MERE CONFIDENCE BACKED BY NOTHING IS MISERY.

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