設常數a>0,(ax2+1x) 4展開式中x3的係數為32,則limn→∞(a+a2+…+an)=( ) A. 14B. 12C. 2D. 1
(ax2+1x)4展開式的通項為Tr+1=a4−rCr4x8−5r2令8−52r=3得r=2展開式中x3的係數為a2C24=32解得a=12∴limn→∞(a+a2+…+an)=limn→∞12(1−(12)n)1−12 =1故選D
(ax2+1x)4展開式的通項為Tr+1=a4−rCr4x8−5r2令8−52r=3得r=2展開式中x3的係數為a2C24=32解得a=12∴limn→∞(a+a2+…+an)=limn→∞12(1−(12)n)1−12 =1故選D