1.

If tn=1/4(n+2)(n+3) for n=1,2,3,... then 1/t1+1/t2+1/t3+........+1/2003

Answer» Tn=1/4(n+2)(n+3)

=>1/Tn=4/((n+2)(n+3))

=>1/Tn=4[1/(n+2)-1/(n+3)

Putting n=1,2,3,4......up to n and summing we get

1/T1+1/T2+1/T3+...Tn=.4(1/3-1/(n+3))

As n-> infinity

We get

1/T1+1/T2+1/T3+.....=4/3

So

1/T1+1/T2+1/T3+.......+.1/2003

=4/3+1/2003

=8015/6009


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