MathematicsMedium64×since 2002Q2966If Sn=∑r=0n1nCr and tn=∑r=0nrnCr, {S_n} = \sum\limits_{r = 0}^n {{1 \over {{}^n{C_r}}}} \,\,and\,\,{t_n} = \sum\limits_{r = 0}^n {{r \over {{}^n{C_r}}},\,}Sn=r=0∑nnCr1andtn=r=0∑nnCrr,then tnSn{{{t_{ n}}} \over {{S_n}}}Sntn is equal toA2n−12{{2n - 1} \over 2}22n−1B12n−1{1 \over 2}n - 121n−1Cn - 1D12n{1 \over 2}n21nCheck answerSkip