(1^2-t_1)+(2^2-t_2)+....+(n^2-t_n)=(n(n^2-1)/(3) ಆಗ t_n ಗೆ ಸಮಾನವಾಗಿರುತ್ತದೆ | 12 | ಪ್ರಗತಿ ಮತ್ತು ಎಸ್...

If (1^2-t_1)+(2^2-t_2)+….+(n^2-t_n)=(n(n^2-1))/(3) then t_n is equal to

If (1^2-t_1)+(2^2-t_2)+......(n^2-t_n)=1/3n(n^2-1) ,then t_n is :(1) n/2 (2) n-1 (3) n+1 (4) n

If (1^(2)-t_(1))+(2^(2)-t_(2))+...+(n^(2)-t_(n))+=(n(n^(2)-1))/(3) then t_(n) is equal to n^(2) b.2nc.n^(2)-2nd .none of these

Let S_n=1/1^2 + 1/2^2 + 1/3^2 +….. + 1/n^2 and T_n=2 -1/n , then :

For sequence (t_(n)), if S_(n)=5(2^(n)-1) then t_(n)= . . .

The sum S_(n) where T_(n)=(-1)^(n)(n^(2)+n+1)/(n!) is

If t_(n)=(1+(-2)^(n))/(n-1) , find t_(6)-t_(5)

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