已知等差数列an的通项公式为an=n^2cosπn
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当n=1时,2S1=a1+1/a1,得a1=1当n=2时,2S2=2(1+a2)=a2+1/a2,得a2=√2-1当n=3时,2S3=2(√2+a3)=a3+1/a3,得a3=√3-√2猜想an=√n
已知等差数列{an}的前三项依次为a-1,a+1,2a+3,故有2(a+1)=a-1+2a+3,解得a=0,故等差数列{an}的前三项依次为-1,1,3,故数列是以-1为首项,以2为公差的等差数列,故
加一个条件,an是正项数列2Sn=(an)²+n-42S(n-1)=[a(n-1)]²+(n-1)-4n>=2则2an=2Sn-2S(n-1)=(an)²-[a(n-1)
设A1=a公差=dAn=a+(n-1)d=a-d+ndA(n+1)=a+ndAnA(n+1)=(a-d+nd)(a+nd)=(nd)^2+(2a-d)nd+a^2+a(a-d)=4n^2-1d^2=4
S1=a1=89,S2=a1+a2=2425,S3的=S2+a3=4849.猜测Sn=(2n+1)2−1(2n+1)2.证明:①当n=1时,由以上可知,猜测成立.②假设n=k时,猜测成立,即SK=(2
由b+1-(b-1)=2b+3-(b+1)得b=0a8=-1公差d=2an=a8+(n-8)*d=2n-17
设等差数列{an}前三项分别为a-d,a,a+d,则由题意得:a−d+a+a+d=−3(a−d)a(a+d)=8,解得:a=−1d=−3或a=−1d=3.当a=-1,d=-3时,首项a1=a-d=-1
A1=a-1A2=a+2A3=2a+3D=(A3-A1)/(3-1)=A-1=(A2-A1)/(2-1)=3所以D=3A1=2AN=2+(N-1)3=3N-1
∵等差数列{a[n]},S[n]=[(a[n]+1)/2]^2∴4S[n]=a[n]^2+2a[n]+1∵4S[n+1]=a[n+1]^2+2a[n+1]+1∴将上面两式相减,得:4a[n+1]=a[
你可以看出公差d=2第一项是A-1所以公式为An=A1+(n-1)d即首项+(n-1)乘以公差d=a-1+(n-1)2=a+2n-3
2a-(a+1)=a+3-2a推导出啊a=2{an}=3,4,5...{an}=a+2(a>=1)
等差数列公式Sn=n(a1+an)/2或Sn=a1*n+n(n-1)d/2注:an=a1+(n-1)d185=a1*10+10*(10-1)d/214=a1+(10-1)d解得a1=5d=3an=5+
∵等差数列{an}的前三项分别为a-1,2a+1,a+7,∴2(2a+1)=a-1+a+7,解得a=2.∴a1=2-1=1,a2=2×2+1=5,a3=2+7=9,∴数列an是以1为首项,4为周期的等
由题:a-1,a+1差为2,所以2a+3-(a+1)=2可得a=0;此等差数列为-1,1,3.通式为:2n-1,n=0,1,2.
设an=a1+(n-1)d=10+(n-1)dSn=na1+(n-1)nd/2=10n+(n-1)nd/2S12=120+66d=-125那么d就算出来了d=-245/66所以an=10+(n-1)(
设an=a1+(n-1)d则a2=a1+da3=a1+2da4=a1+3da7=a1+6d因为等差数列{an}的前四项和为10所以,a1+a2+a3+a4=10即4a1+6d=10.①又因a2,a3,
1/A(n+1)=(An+2)/(2An)=1/An+1/2{1/An}是公差为1/2的等差数列1/A1=1/21/An=n/2An=2/n
半天你也没打对应该是an=16-3n求{|an|}的前n项和公式an=16-3n>0得n≤5∴n≤5时,|an|=anSn=(a1+an)*n/2=(13+16-3n)*n/2=(29-3n)n/2=
a2=a1+da4=a1+3da2^2=a1a4a2^2=(a1+d)^2=a1^2+2a1d+d^2a1a4=a1(a1+3d)=a1^2+3a1da1^2+2a1d+d^2=a1^2+3a1da1
Sn、an、1成等差,则2an=Sn+1(n=1时,得a1=1),当n≥2时,有2a(n-1)=S(n-1)+1,则2an-2a(n-1)=an,即an/[a(n-1)]=2=常数,所以{an}是等比