Question

已知等差数列｛an｝满足a2=0.a6+a8=-10　,求数列an/2的n-1次方的前n项和

Answer 1

s=1＋0-1／2�0�5-2／2�0�6＋……－﹙n－2﹚／[2^﹙n-1﹚] �0�5s=�0�5＋0－1／2�0�6＋……－﹙n－1﹚／[2^﹙n-1﹚]－﹙n－2﹚／[2^﹙n﹚] s-�0�5s=�0�5-1／2�0�5－1／2�0�6＋……－1／[2^﹙n-1﹚]＋﹙n－2﹚／[2^﹙n﹚] �0�5s=�0�5-﹛1／2�0�5＋1／2�0�6＋……＋1／[2^﹙n-1﹚]﹜＋﹙n－2﹚／[2^﹙n﹚] =�0�5－﹛1／2�0�5－1／[2^﹙n﹚]﹜／﹙1-�0�5﹚＋﹙n－2﹚／[2^﹙n﹚] =�0�5－1／2＋2／[2^﹙n﹚]＋﹙n－2﹚／[2^﹙n﹚] =n／[2^﹙n﹚] ∴s=n／[2^﹙n-1﹚]