等差数列中sm=n,sn=m则sm n=-(m n)

Sm=a1m+m(m-1)d/2=nSn=a1n+n(n-1)d/2=ma1m+m(m-1)d/2-[a1n+n(n-1)d/2]=n-ma1(m-n)+(m+n-1)(m-n)d/2=n-ma1+(

Sn=(a1+an)n/2=(a1+a1+nd-d)n/2=n(a1-d/2)+dn²/2=an²+bn其中,a=(a1-d/2),b=d/2这么设是为了表示的方便而已.所以有,S

Sm=a1m+m(m-1)d/2=nSn=a1n+n(n-1)d/2=m相减a1m+m(m-1)d/2-[a1n+n(n-1)d/2]=n-ma1(m-n)+(m+n-1)(m-n)d/2=n-m两边

Sm+n=-(m+n)Sn=na1+1/2n(n-1)*d=n^2/2*d+(a1-1/2d)n所以可将Sn表示成An^2+Bn表示,即Sn=An^2+Bn则由题意有Sm=n=Am^2+BmSn=m=

Sm=a1m+m(m-1)d/2=n(1)Sn=a1n+n(n-1)d/2=m(2)(1)-(2)a1(m-n)+(m+n-1)(m-n)d/2=n-ma1+(m+n-1)d/2=-1a1=-1-(m

运用简单的等差数列的性质和均值不等式即可证到!显然：当m不等于n时Sn=n/m,Sm=m/n;两式相减,可得公差：d=2/mn;Sm+n=Sm+Sn+mnd=m/n+n/m+2>4;Sm+n>4当m=

1.∵S[n]=m,S[m]=n (m＞n)∴S[n]=na[1]+n(n-1)d/2=m 【1】 S[m]=ma[1]+m(m-1)d/2=n 【2】由【1】

原问题即：有两个数列,{An}{Bn},若Sm:Sn=m^2:n^2求Am:Bn(公差分别为d1,d2)Sm=A1d+0.5*m(m-1)d1=0.5d1m^2+m(A1-0.5d1)Sn=0.5d2

Sm=Snma1+m(m-1)d/2=na1+n(n-1)d/2(m-n)a1+(m²-m-n²+n)d/2=0(m-n)a1+[(m+n)(m-n)-(m-n)]d/2=0a1(

Sn=A1×n+n(n-1)d/2=(d/2)×n^2+(A1-d/2)×n题目中A=d/2B=A1-d/2Sn=A×n^2+B×n=Sm=A×m^2+B×mA×m^2-A×n^2+B×m-B×nA×

设等差数列首项为a1,公差为d由Sm=m*a1+m*(m-1)*d/2=m/nSn=n*a1+n*(n-1)*d/2=n/m两式相减得a1=1/n+1/m-d*(m+n-1)/2∴Sm+n=(m+n)

设An=kn+bSn=n(kn+k+2b)/2=mSm=Sn(kn+k+2b)/2=[n(kn+k+2b)/2]{[kn(kn+k+2b)/2]+l+2b}/2=nS(m+n)={[n(kn+k+2b

由题意可设Sn=pn2+qn，则Sn=pn2+qn=m，①Sm=pm2+qm=n    ②①-②得：p（n2-m2）+q（n-m）=m-n，即p（m+n）+q=

证明：由数列为等差数列,可设其前n项和Sn=An^2+BnSm=Am^2+Bm=p,（1）Sp=Ap^2+Bp=m（2）（1）+（2）得A(m^2+p^2)+B(m+p)=m+pp*(1)-m*(2)

证：设公差为dSm=Spma1+m(m-1)d/2=pa1+p(p-1)d/2(m-p)a1+[m(m-1)-p(p-1)]d/2=0(m-p)a1+[(m²-p²)-(m-p)]

由于{an}为等差数列则:设an=a+nd,d为公差则有:Sm=am+dm(m+1)/2=nSn=an+dn(n+1)/2=m解得:d=-(2m+2n)/mna=(m^2+n^2+mn+m+n)/mn

没有这样的结果正项等比数列,每一项都是正的除非m=n不然,前m项的和加上几个正数怎么可能与前n项的和相等再问：不好意思，打错了，是｛bn｝的前n项积Tn再答：b1b1qb1q²。。。b1q^

依题意可知ma1+m(m−1)d2=na1+n(n−1)d2，整理得a1+n+m−12d=0∴Sm+n=（m+n）a1+(n+m−1)(n+m)2d=（n+m）（a1+n+m−12d）=0故答案为：0

Sm=ma1＋(1/2)m(m－1)dSn=na1＋(1/2)n(n－1)d两式相减,得：(m－n)a1＋(1/2)d×[m(m－1)－n(n－1)](m－n)a1＋(1/2)d×[(m²－

0等差数列的前N项和是过原点的二次函数因为Sm=Sn=L所以该函数的对称轴是X=(M+N)/2所以原点关于该轴的对称点是Sm+n即Sm+n=0又Sm+n=(m+n)(a1+a(m+n))/2所以a1+