The sum of the series `1/(1!(n-1)!)+1/(3!(n-3)!)+1/(5!(n-5)!)+…..+1/((n-1)!1!)` is = (A) `1/(n!2^n)` (B) `2^n/n!` (C) `2^(n-1)/n!` (D) `1/(n!2^(n-1)`

Sum of the series S_(n) =(n) (n) + (n-1) (n+1) + (n-2) (n+2) + …+ 1(2n-1) is

Find the sum of the series: 1.n+2.(n-1)+3.(n-2)+...+(n-1).2+n.1

Find the sum of the series 1xx n+2(n-1)+3xx(n-2)+...+(n-1)xx2+n xx1

The sum of the series 1+(1)/(2) ""^(n) C_1 + (1)/(3) ""^(n) C_(2) + ….+ (1)/(n+1) ""^(n) C_(n) is equal to

The sum of the series: (1)/(log_(2)4)+(1)/(log_(4)4)+(1)/(log_(8)4)+...+(1)/(log_(2n)4) is (n(n+1))/(2) (b) (n(n+1)(2n+1))/(12) (c) (n(n+1))/(4) (d) none of these

The arithmetic mean of the series 1,2,4,8,16,....2^(n) is (2^(n)-1)/(n+1) (b) (2^(n)+1)/(n) (c) (2^(n)-1)/(n) (d) (2^(n+1)-1)/(n+1)

The sum of n terms of the series 5/1.2.1/3+7/2.3.1/3^2+9/3.4.1/3^3+11/4.5.1/3^4+.. is (A) 1+1/2^(n-1).1/3^n (B) 1+1/(n+1).1/3^n (C) 1-1/(n+1).1/3^n (D) 1+1/2n-1.1/3^n

Find the sum of n terms of the series (1)/(2*4)+(1)/(4*6)+... (A) (n)/(n+1) (B) (n)/(4(n+1)) (C) (1)/((2n)(2n+2))( D) )(1)/(2^(n)(2^(n)+2))