IIT JEE , AIEEE Forum - Mathematics , Algebra

shajikoshy

Find the value of S.

rohith291991

answer is (1999)(2001)/2000 basically just find the lcm...numerator will come out to be a perfect square...consider the nth term ... we see that if n=1 numerator becomes 3(after finding sqr root..) and then if n=2 numerator is 7...and if n=3 numerator is 13 etc...here u see that the numerators form an ap in which the common diff is increasing...so u cud go by the conventional way and find the expression for the numerator....but instead u can easily see that the nth numerator is given by...n²+n+1...hence the general nth term is (n²+n+1)/n(n+1) = 1+1/n(n+1)now we have to find _[n=1]

^{[n=1999 ]} [1+1/n(n+1)] but 1/n(n+1)=1/n-1/(n+1) this becomes
1999+ _[n=1]

^{[n=1999 ]} 1/n-1/(n+1) now all the terms get canceled except 1 and -1/2000 hence the final answer is 1999+1-1/2000 =1999+1999/2000=(1999)(2001)/2000== 3999999/2000

bvsatyaram

Well done Rohith....

harsha_27

brilliant answer rohith..............

imsimple

dude i dint understand how 1/n{n+1}=1/n-1/n+1

imsimple

k i got it

shajikoshy

great work.

Thank you for the solution..

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