若Sm=p,Sp=m(m≠p),则Sm p

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

Sn=[(a1+a1+(n-1)d]*n/2=[2a1+(n-1)d)]*n/2Sm/m={[2a1+(m-1)d)]*m/2}/m=a1+(m-1)d/2Sn/n=a1+(n-1)d/2Sp/p=a

Sm+p=a1+a2+…＋am+am+1+…+am+p=Sm+（a1+md）+（a2+md）+…＋（ap+md）=Sm+Sp+mpd=m+p+mpd=m+p+2mp(m/p-p/m)/(p-m)d/2

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】

则m(1n−1p)+n(1m−1p)−p(1m+1n)=mn-mp+nm-np-pm-pn=m−pn+n−pm-m+np由题意可得：m-p=-n，m-p=-n，n-p=-m，m+n=p，∴可得：m(1

依题意得：n=12m，p=2m，∴m+n+p=m+12m+2m=72m．故选C．

等差数列中,若Sm=Sn,m≠n,则S(m+n)=0证明：设等差数列{an}的首项为a1,公差为dS(n)=na1+n(n-1)d/2所以ma1+m(m-1)d/2=na1+n(n-1)d/2故(m-

(1)在等差数列{an}中,若m+n=p+q＜==＞am+an=ap+aq(2)等差数列{an}中,d/2=(Sn/n-Sm/m)/(n-m)(3)数列{an}是等差数列＜==＞Sk=Ak+Bk(4)

证明:因为在等差数列中m＋n=p＋q,所以am＋an=ap＋aq,所以m*am＋n*an=p*ap＋q*aq,m*(a1＋am)＋n*(a1＋am)=p*(a1＋ap)＋q(a1＋aq),所以m*(a

S(x)=x(x+1)/2p=n(n+1)/m(m+1)n^2+n=pm(m+1)(2n+1)^2=p(2m+1)^2-p+1设u=2n+1v=2m+1那么u^2-pv^2=1-p显然这个方程存在解u

等差数列中na1+n(n-1)d/2=-dn²/2+(a1+d/2)n,∴可设Sn=An²+Bn.1.（Sp-Sq）/(p-q)=(Ap²+Bp-Aq²-Bq)

设Sn=a*n^2+b*nSk=SmSp=0根据二次函数对称性k+m=p

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)

PrivateSubCommand1_Click()Fori=1To2000IfPrime(Sum_Y(i))=TrueAndPrime(i)=FalseThenPrintiNextEndSub'求素

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)]

a(n)=a+(n-1)d.s(n)=na+n(n-1)d/2.p=s(m)=ma+m(m-1)d/2.p^2=mpa+mp(m-1)d/2.m=s(p)=pa+p(p-1)d/2.m^2=mpa+m

由Sn=na1+n(n-1)d/2=dn^2+(a1-d/2)n,1.当d不等于0时函数为一元二次方程,且恒过定点（0,0）,2.当d=0时函数Sn=n*a1.显然对2来说,是一条直线不可能满足Sm=

(m+n)(p+q)-(m+n)(p-q)=（m+n)(p+q-P+q)=(m+n)×2q=2q(m+n)

5n再问：为什么？再答：带进去

等于P.因为sm=sp,但是m不等于p,所以s=0.sm+p=p