IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry - iit-2008

VMC

Find the sum of the series:

1/3.5+1/5.7+1/7.9+----------------------- upto infinity

(A) 1/6 (B) 1/3

(C) 1/2 (D) 5/6

pls also explain ur answer.pls also tell me how to start thinking in such types of problems.

iitkgp_bipin

1/3.5 = (1/2)(1/3 - 1/5)

1/5.7 = (1/2)(1/5 - 1/7)

1/7.9 = (1/2)(1/7 - 1/9)

Hence sum = (1/2)(1/3 - 1/5) + (1/2)(1/5 - 1/7) + (1/2)(1/7 - 1/9) + .......

= (1/2)(1/3 - 1/5 + 1/5 - 1/7 + 1/7 -1/9 +............)

In this way the terms get cancel out and the remaining term will be (1/2)(1/3)

Hence sum = 1/6

ramyadiamond

In these types of questions, if u dont know how to break up the terms, then first write the n^th term. here, the term would be,

t_n= 1/(2n+1)(2n+3) for n=1,2,3,...........infinity.

now, put the sigma sign for this term, to get the sum of the series, for values of n varying from 1 to infinity.

_{[ n=1]}^{[infinity ]} t_n= _{[ n=1]}^{[infinity ]} 1/(2n+1)(2n+3)

1/(2n+1)(2n+3) can b written as 1/2(2n+1) -1/2(2n+3) = 1/2[1/2n+1 -1/2n+3 ] using the concept of partial fractions.

So, as Bipin sir has further solved it above, when u put different values of n, u'll get similar types of values, whic can get cancelled. Here,

1/2{1/3-1/5 +1/5 -1/7........................+upto infinity}

So, u can see, that except the first term, all the other terms will get cancelled as the terms strtch to infinity.

so, sum becomes equal to = 1/2 *1/3 =1/6

Hope this helps.

Thank you.

catch_arnnie

the method talked here is usually known as 'V^_nmethod'....

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