if `S_n=1/(3!)+5/(4!)+11/(5!)+19/(6!)+..................t_n` then:

If S_(1), S_(2), S_(3),….., S_(n) are the sum of infinite geometric series whose first terms are 1,3,5…., (2n-1) and whose common rations are 2/3, 2/5,…., (2)/(2n +1) respectively, then {(1)/(S_(1) S_(2)S_(3))+ (1)/(S_(2) S_(3) S_(4))+ (1)/(S_(3) S_(4)S_(5))+ ........."upon infinite terms"}=

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

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 (1)/(""^(4)C_(n))=(1)/(""^(5)C_(n))+(1)/(""^(6)C_(n)) , then value of n is:

Sum of n terms of the series 1/(1.2.3.4.)+1/(2.3.4.5) +1/(3.4.5.6)+....

Sum of n terms of the series 1/(1.2.3.4.)+1/(2.3.4.5) +1/(3.4.5.6)+....

If S_n=sum_(r=0)^n 1/(nC_r) and t_n=sum_(r=0)^n r/(nC_r), then t_n/S_n=

Prove the following by the principle of mathematical induction: 1/(3. 7)+1/(7. 11)+1/(11. 15)+.....+1/((4n-1)(4n+3))=n/(3(4n+3))

Let S_n(n leq 1) be a sequence of sets defined by S_1{0},S_2={3/2,5/2},S_3={15/4,19/4,23/4,27/4},....... then

In an A.P. if S_1=T_1+T_2+T_3+.....+T_n (n is odd) S_2=T_2+T_4+T_6+.........+T_(n-1) , then find the value of S_1//S_2 in terms of ndot