Consider `S_n=1/(3^2+1)+1/(4^2+2)+1/(5^2+3)+1/(6^2+4)+,,,` upto n terms then

1^(2) + 3^(2) + 5^(2) + …. upto n terms =

If S_(n)=(1.2)/(3!)+(2.2^(2))/(4!)+(3.2^(3))/(5!)+... upto n terms then the sum infinite terms is

S_n=3/(1^2)+5/(1^2+2^2)+7/(1^2+2^2+3^2)+ upto n terms, then a. S_n=(6n)/(n+1) b. S_n=(6(n+2))/(n+1) c. S_oo=6 d. S_oo=1

Find 1^2/1 + (1^2 + 2^2) / 2 + (1^2 + 2^2 + 3^2) / 3 + …..upto n terms.

1+(1+2)/(2)+(1+2+3)/(3) +(1+2+3+4)/(4)+.. . to n terms =

If S_(n)=3+(1+3+3^(2))/(3!)+(1+3+3^(2)+3^(3))/(4!) ………… upto n -terms Then the value of [lim_(ntooo)S_(n)] is (where [.] represent G.I.F)

(1-1/n) + (1 - 2/n) + (1 - 3/n)+ …… upto n terms = ?

Prove by the method of induction (1)/(1*2)+(1)/(2*3)+(1)/(3*4)+………… upto n terms =(n)/(n+1) where n in N

The odd value of n for which 704 + 1/2 (704) + 1/4 (704) + ... upto n terms = 1984 - 1/2 (1984) + 1/4 ( 1984) - .... upto n terms is :