a2 a ( )=a c(a 1不等于0)
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把3个式子统一起来,写成矩阵形式:A*[a1a2a3]=[a1a2a3]*110011001记P=[a1a2a3],J=110011001(其实J就是一个特征值为1的三阶Jondan块).则有AP=P
a4=a1+3d,a8=a1+7da1,a4,a8成等比数列,a4²=a1×a8(a1+3d)²=a1×(a1+7d)整理得,a1=9d所以an=a1+(n-1)d=(n+8)d﹙
这是我先在word里写再截过来的,不清楚或者不懂的话再问我再问:(1)证,当n=1时,a1=a1q^o=a1,成立。假设当n=k时等式成立,即ak=a1q^k-1,那么当n=k+1时,ak+1=akq
a5=a1+4d,a17=a1+16d因为a1,a5,a17成等比数列所以(a1+4d)^2=a1*(a1+16d)故(a1)^2+8a1*d+16d^2=(a1)^2+16a1*d即2d^2=a1*
a3=(a1+a5)/2b3=√b1b5=√a1a5∵a1≠a3,∴d≠0,a1≠a5a3-b3=(a1+a5-2√a1a5)/2=(√a1-√a5)²/2>0∴a3>b3再问:懂了
a1+d=a2=b1*q=b2b1q-a1-d=0b1=a1不等于a2,则q不等于1a1=d/(q-1)因an>0,则d>0,否则,总有an小于0的时候.0b3a4=a3+db4=b3*qa4>a3b
这个没法求得,这个“矩阵”是1x1的,如果a1b1+a2b2+a3b3=0,则秩为0,否则为1再问:能不能给我写下过程啊谢谢你了再答:哪步需要过程?你按照矩阵乘法乘一下不就得到答案了?
因为a1,a2,a5成等比数列,根据等比中项公式:a2^2=a1xa5(1+d)^2=1x(1+4d)d^+2d+1=4d+1d^2-2d=0d=0或d=2因为d不等于0,所以d=2
因为1/anan+1=1/an*(an+d)=1/d[1/an-1/(an+d)]=1/d[1/an-1/an+1]所以1/a1a2+1/a2a3+…+1/anan+1=1/d[1/a1-1/a2+1
B2=IF(A1=A2,B1,0)或B2=IF(A1=A2,B1,"")
A4输入公式=if(and(a10,a20,a30),A5,"")如果是并列条件=if(or(a10,a20,a30),A5,"")再问:如果 A1=0 且 A2=0&n
解析:∵a1=4,a7=4+6d,a10=4+9d∴a7^2=a1*a10,即(4+6d)^2=4(4+9d)∵d≠0∴d=-1/3即a1=4,a7=2,a10=1∴q=a2/a1=1/2∴Sn=4*
原式=a2a-1-(a+1)=a2a-1-(a+1)(a-1)a-1=a2-a2+1a-1=1a-1,故答案为:1a-1.
是A1,A3,A4等比数列吧?∵A1,A3,A4等比数列∴(a3)²=(a1)×(a4)(a1+2d)²=(a1)(a1+3d)a²₁+4d²+4a
Sn=nan-2n(n-1)Sn=n(Sn-S(n-1))-2n(n-1)(n-1)Sn-nS(n-1)=2n(n-1)Sn/n-S(n-1)/(n-1)=2Sn/n-S1/1=2(n-1)Sn/n=
a(n)=1+(n-1)da(n+1)=1+ndSn=(1+an)n/2=(2+nd-d)n/2(1+Sn)/(n(1-a(n+1)))=-((4+nd-d)/n)/(2n(nd))=-2/(nd)-