(n-1)π nπ an=sinx x^p的积分,研究收敛性
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1.an-an-1=2(n-1)-1=2(n-1)2n-2=-12n=2-12n=1n=1/22.3+(n-1)(-2)=-2n-53-2n+2=-2n-55=-5题目有错,无解.3.2+(n-1)x
(1)bn=a(2n+1)+4n-2b(n+1)=a(2n+3)+4(n+1)-2=a(2n+2+1)+4n+2=a(2n+2)-2(2n+2)+4n+2=a(2n+1+1)-2(2n+2)+4n+2
试试再答:再答:再答:再答:搞定。
(1)由已知a2=2a1+2,a3=2a2+3=4a1+7,若{an}是等差数列,则2a2=a1+a3,即4a1+4=5a1+7,得a1=-3,a2=-4,故d=-1. &nbs
(Ⅰ)∵a1=-58,an+1-an=1n(n+1),∴a2=−18,a3=124  
不知道你的题目是不是这样
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之前你出过这种题了吧,原来让求的是前30项.也不说清楚是从a0还是a1开始,不过不要紧a0=0;之前求的是S29,S30如下cos(nπ/3)^2-sin(nπ/3)^2=1-2sin(nπ/3)^2