Find the sum to `n` terms of the series `1/(1+1^2+1^4)+2/(1+2^2+2^4)+3/(1+3^2+3^4)+..........` that means `t_r = r/(r^4+r^2+1)` find ` sum_(r=1)^n`

Find the sum of n terms \ of the series: 1/(1. 2)+1/(2. 3)+1/(3. 4)+...............+1/(n.(n+1))

Find the sum of n terms \ of the series: 1/(1. 2)+1/(2. 3)+1/(3. 4)+dot+1/(ndot(n+1))

If T_r=r(r^2-1), then find sum_(r=2)^oo1/Tdot

Sum of the series sum_(r=1)^(n) (r^(2)+1)r! is

Find the sum of firs 100 terms of the series whose general term is given by T_(r)=(r^(2)+1)r! .

Sum the following series to n terms and to infinity : (i) (1)/(1.4.7) + (1)/(4.7.10) + (1)/(7.10.13) + ... (ii) sum_(r=1)^(n) r (r + 1) (r + 2) (r + 3) (iii) sum_(r=1)^(n) (1)/(4r^(2) - 1)

sum_(r=1)^10 r/(1-3r^2+r^4)

Sum of the series 1+(1+2)+(1+2+3)+(1+2+3+4)+... to n terms be 1/m n(n+1)[(2n+1)/k+1] then find the value of k+m

Assertion: If n is odd then the sum of n terms of the series 1^2+2xx2^2+3^2+2xx4^2+5^2+2xx6^2+7^2+…is (n^2(n+1))/2. If n is even then the sum of n terms of the series. 1^2+2xx2^2+3^2+2xx4^2+5^2+2xx6^2+…..is (n(n+1)^2)/2 (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true R is not te correct explanation of A (C) A is true but R is false. (D) A is false but R is true.

Find the sum sum_(r =1)^(oo) tan^(-1) ((2(2r -1))/(4 + r^(2) (r^(2) -2r + 1)))