how to solve this sort of questions

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

imhimanshujaggi's picture
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Last seen: 11 years 5 weeks ago
Joined: 2007-06-18
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
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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???

neerej (not verified)
Incognito's picture
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

rajorshi's picture
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Last seen: 10 years 21 weeks ago
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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....!

imhimanshujaggi's picture
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Last seen: 11 years 5 weeks ago
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not suggestable...

cum on rajorshi...!!!!

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

searchin-life's picture
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Last seen: 11 years 28 weeks ago
Joined: 2007-03-03
You must have mathematics in

You must have mathematics in your graduation level that’s why in the first look you can determine what to write in place 1/1.2

For beginner’s rajorshi's solutions given a good explanation and to start with they shd solve like that only.

rajorshi's picture
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Last seen: 10 years 21 weeks ago
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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.

Incognito
Incognito's picture
thank you

super way of solving i appreatiate you

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