The sum `(3xx1^3)/1^2+ (5xx(1^3+2^3))/(1^2+2^2)+...=`

The sum (3 xx 1^(3))/(1^(2)) + (5 xx (1^(3) + 2^(3)))/(1^(2) + 2^(2)) + (7 xx (1^(3) + 2^(3) + 3^(3)))/(1^(2) + 2^(2) + 3^(2)) + ....

The sum (3 xx 1^(3))/(1^(2)) + (5 xx (1^(3) + 2^(3)))/(1^(2) + 2^(2)) + (7 xx (1^(3) + 2^(3) + 3^(3)))/(1^(2) +2^(2) + 3^(2))... upto 10th term is

The sum up to 10 terms of the series (3 xx 1^(3))/(1^(2)) + (5 xx (1^(3) + 2^(3)))/(1^(2) + 2^(2)) + (7 xx (1^(3) + 2^(3) + 3^(3)))/(1^(2) + 2^(2) + 3^(2)) + ....

Finds the sum of first n terms of the series 3/(1^2xx2^2)+5/(2^2xx3^2)+7/(3^2xx4^2)+...... and hence deduce the sum of infinity.

Find the sum (3)/(1xx2)xx(1)/(2)+(4)/(2xx3)xx((1)/(2))^(2)+(5)/(3xx4)xx((1)/(2))^(2)+... to n terms.

Find the sum (1^4)/(1xx3)+(2^4)/(3xx5)+(3^4)/(5xx7)+......+(n^4)/((2n-1)(2n+1))

Find the sum (1^4)/(1xx3)+(2^4)/(3xx5)+(3^4)/(5xx7)+......+(n^4)/((2n-1)(2n+1))

Show that (1xx2^(2)+2xx3^(2)+...+n xx(n+1)^(2))/(1^(2)xx2+2^(2)xx3+...+n^(2)xx(n+1))=(3n+5)/(3n+1)

Find the sum 3/(1xx2)xx1/2+4/(2xx3)xx(1/2)^2+5/(3xx4)xx(1/2)^3+ to n terms.

Find the sum 3/(1xx2)xx1/2+4/(2xx3)xx(1/2)^2+5/(3xx4)xx(1/2)^2+ ton terms.