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# If $\;p=1+\frac{1}{2}+\frac{1}{3}+\;....\;+\frac{1}{n}\;$ then $\;S=\frac{1^2}{1^3}+\frac{1^2+2^2}{1^3+2^3}+\large\frac{1^2+2^2+3^2}{1^3+2^3+3^3}+\;....\;$ upto n terms equal to

$(a)\;\frac{4}{3}\;p-1\qquad(b)\;\frac{4}{3}\;p+\frac{1}{n}\qquad(c)\;\frac{4}{3}\;p\qquad(d)\;\frac{4}{3}-\frac{2}{3}\;\frac{n}{n+1}$

Can you answer this question?