If `T_n=(n^2+1)n ! & S_n=T_1+T_2+T_3+....+T_n, l e t I_10/S_10=a/b` where a & b are relatively prime

If t_n=sum_1^n n , find S_n=sum_1^n t_n .

If t_n = 3^(n+1) , then S_6 - S_5 = __________.

If in an AP, t_1 = log_10 a, t_(n+1) = log_10 b and t_(2n+1) = log_10 c then a, b, c are in

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

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

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))=

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))=

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))=

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))=

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))=