If the sum of the series `sum_(n=0)^oor^n ,|r|<1` is `s ,` then find the sum of the series `sum_(n=0)^oor^(2n)` .

Find the sum of the series sum_(r=1)^n rxxr !dot

Find the sum of the series (sum_(r=1)^(n) rxxr !)

The sum of the series sum_(r=0) ^(n) "r."^(2n)C_(r), is

The sum of the series sum_(r=0) ^(n) ""^(2n)C_(r), is

The sum of the series sum_(r=1)^(n) (-1)^(r-1).""^(n)C_(r)(a-r), n gt 1 is equal to :

In a geometric series, the first term is a and common ratio is rdot If S_n denotes the sum of the terms and U_n=sum_(n=1)^n S_n ,then r S_n+(1-r)U_n equals (a) 0 b. n c. n a d. n a r

Sum of the series sum_(r=0)^n (-1)^r ^nC_r[i^(5r)+i^(6r)+i^(7r)+i^(8r)] is

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

The sum of series Sigma_(r=0)^(r) (-1)^r(n+2r)^2 (where n is even) is

If sum_(r=1)^(n)T_(r)=(n(n+1)(n+2)(n+3))/(12) where T_(r) denotes the rth term of the series. Find lim_(nto oo) sum_(r=1)^(n)(1)/(T_(r)) .