and also of responses recorded in the intact cortex and in an isolated strip of cortex, suggests that type E1 responses are due mainly to the intracortical spread of excitation, whereas responses of types E2 and E3 are due to an excitatory inflow into the cortex from subcortical formations....
(Ⅱ)由(Ⅰ)可知an=6-4n,求得Sn= n(2+6-4n) 2 =4n-2n2.再由Sk=-48,可得4k-2k2=-48,解得k的值. 解答:解:(Ⅰ)设等差数列{an}的公差为d,则an=a1+(n-1)d. 由a1=2,a3=-6,可得2+2d=-6,解得d=-4. 从而,an=2+(n-1)×(-4)=6-4n.---(5分) (Ⅱ)由...
解答一 举报 你是想说''sk,s2k-sk,s3k-s2k.一定是?''吧,仍为等比数列,且公比为原公比的k次方. 解析看不懂?免费查看同类题视频解析查看解答 相似问题 等比数列{an}的一次每k项之和所构成的数列;sk,s2k-sk,s3k-s2k···一定是什么数列? 用S表示等差数列{an}的前k项和求证:Sk、S2k-Sk、S3k-S2...
(2)设Sk=a1+a3+…+a2k-1,Tk=a2+a4+…+a2k,,求使Wk>1的所有k的值,并说明理由。 试题答案 解:(1)因为所以一般地,当时,即所以数列是首项为0、公差为4的等差数列,因此当时,所以数列{a2k}是首项为2、公比为2的等比数列,因此故数列{an}的通项公式为...