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If tn=1/4(n+2)(n+3) for n=1, 2 ,3,.... ...

If `t_n=1/4(n+2)(n+3)` for `n=1, 2 ,3,....` then `1/t_1+1/t_2+1/t_3+....+1/(t_(2003))=`

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If t_(n)=(1)/(4)(n+2)(n+3) , n=1,2,3, .. then (1)/(t_(1))+(1)/(t_(2))+(1)/(t_(3))+..+(1)/(t_(2003))=

Suppose a series of n terms given by S_(n)=t_(1)+t_(2)+t_(3)+ . . . . +t_(n) then S_(n-1)=t_(1)+t_(2)+t_(3)+ . . . . +t_(n-1),nge1 subtracting we get S_(n)-S_(n-1)=t_(n),nge2 surther if we put n=1 is the first sum then S_(1)=t_(1) thus w can write t_(n)=S_(n)-S_(n-1),nge2 and t_(1)=S_(1) Q. The sum of n terms of a series is a.2^(n)-b . where a and b are constant then the series is

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