题目：设S= \frac{2}{1× 3× 5}+ \frac{{2}^{2}}{3× 5× 7}+ \frac{{2}^{3}}{5× 7× 9}+…+ \frac{{2}^{48}}{95× 97× 99}，T= \frac{1}{1× 3}+ \frac{2}{3× 5}+ \frac{{2}^{2}}{5× 7}+…+ \frac{{2}^{47}}{95× 97}，则12S-3T= ( ).
选项：
bullet: A
text: 1- \frac{{2}^{48}}{9603}
bullet: B
text: \frac{{2}^{48}}{9603}
bullet: C
text: 1- \frac{{2}^{48}}{3201}
bullet: D
text: \frac{{2}^{48}}{3201}
答案：C
解析：解:∵S= \frac{2}{1× 3× 5}+ \frac{{2}^{2}}{3× 5× 7}+ \frac{{2}^{3}}{5× 7× 9}+…+ \frac{{2}^{48}}{95× 97× 99}，T= \frac{1}{1× 3}+ \frac{2}{3× 5}+ \frac{{2}^{2}}{5× 7}+…+ \frac{{2}^{47}}{95× 97}，则， ∴4S=2( \frac{1}{3× 5}- \frac{1}{3× 5})+2^{2}( \frac{1}{3× 5}- \frac{1}{5× 7})+2^{3}( \frac{1}{5× 7}- \frac{1}{7× 9})+…+2^{48}( \frac{1}{95× 97}- \frac{1}{97× 99}) = \frac{2}{1× 3}+ \frac{2}{3× 5}+ \frac{4}{5× 7}+ \frac{8}{7× 9}+…+ \frac{{2}^{47}}{95× 97}+ \frac{{2}^{48}}{97× 99}， ∴4S-T= \frac{1}{1× 3}- \frac{{2}^{48}}{97× 99}， ∴12S-3T=3( \frac{1}{1× 3}- \frac{{2}^{48}}{97× 99})=1- \frac{{2}^{48}}{97× 33}=1- \frac{{2}^{48}}{3201}， 故选:C.