The sum to `50` terms of series `(3)/(1^(2))+(5)/(1^(2)+2^(2))+(7)/(1^(2)+2^(2)+3^(2))+......` is equal to

The sum to n terms of the series (3)/(1^(2))+(5)/(1^(2)+2^(2))+(7)/(1^(2)+2^(2)+3^(2))+--

The sum of the series (3)/(1^(2))+(5)/(1^(2)+2^(2))+(7)/(1^(2)+2^(2)+3^(2))+... is

Sum of the n terms of the series (3)/(1^(2))+(5)/(1^(2)+2^(2))+(7)/(1^(2)+2^(2)+3^(3))+"......." is

The sum of the series (3)/(1^(2))+(5)/(1^(2)+2^(2))+(7)/(1^(2)+2^(2)+3^(2))+...... upto n terms,is

The sum of 50 terms of the series (3)/(1^(2))+(5)/(1^(2)+2^(2))+(7)/(1^(2)+2^(2)+3^(2))+... is (100)/(17) b.(150)/(17)c(200)/(51) d.(50)/(17)

the sum (3)/(1^(2))+(5)/(1^(2)+2^(2))+(7)/(1^(2)+2^(2)+3^(2))+...... upto 11 terms

The sum to 50 terms of the series 3/1^2+5/(1^2+2^2)+7/(1^+2^2+3^2)+….+… is

The sum of 10 terms of the series (3)/(1^(2)xx2^(2))+(5)/(2^(2)xx3^(2))+(7)/(3^(2)xx4^(2))+… is

Find the sum to n terms of the series: (3)/(1^(2).2^(2))+(5)/(2^(2).3^(2))+(7)/(3^(2).4^(2))+

Sum of first 20 terms of (3)/(1^(2))+(5)/(1^(2)+2^(2))+(7)/(1^(2)+2^(2)+3^(2))+... upto 20 terms is :