若m不等于n,两个等差数列
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既然是等差数列Sn=na1+n(n-1)d/2=mSm=ma1+m(m-1)d/2=n上式相减得(n-m)a1+(n^2-n-m^2+m)d/2=m-na1+(n+m-1)d/2=-1S(n+m)=(
设公差为d.Sn=Smna1+n(n-1)d/2=ma1+m(m-1)d/2(m-n)a1+(m²-n²-m+n)d/2=0(m-n)a1+[(m+n)(m-n)-(m-n)]d/
设m>n,则sm-sn=am+a(m-1)+……+a(n+1)=0得am+a(n+1)=0,因为p+q=m+n则am+an=ap+aq所以am+a(n+1)=a1+a(m+n)得a1+a(m+n)=0
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=
设Sn=pn^2+qnpm^2+qm=npn^2+qn=mp(m^2-n^2)=(q+1)(n-m)-p(m+n)=q+1所以S(m+n)=p(m+n)^2+q(m+n)=(m+n)[p(m+n)+q
1、设公差为d,则n=m+(m-n)d=>d=-1所以am+n=am+nd=02、x1-x2=(b-a)/3y3-y1=2(b-a)/4=(b-a)/2所以(y3-y1)/(x2-x1)=3/2
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(
设首项为a1,公差为d,Sn=na1+n*(n-1)d/2=m,Sm=ma1+m*(m-1)d/2=n,两式相减,得(n-m)a1+[(n-m)(n+m)-(n-m)]d/2=-(n-m)a1+[n+
B(m+n)=(nBn/mBm)开(n-m)次方根
把Sn=m2与Sm=n2的式子列出来,两式相减,得的式子有公因式(n-m),消去它,得的式f子再整体乘上(m+n),左边式子就是S(m+n)的展开式,右面是答案:-(m+n)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
am=a1+(m-1)d=n.(1)an=a1+(n-1)d=m.(2)(2)-(1)得m-n=(n-m)d所以d=-1代回(1)式得a1+1-m=n所以a1=m+n-1所以a(m+n)=a1+(m+
Sm=(a1+am)*m/2Sn=(a1+an)*n/2所以(a1+am)*m/2=(a1+an)*n/22ma1+m(m-1)d=2na1+n(n-1)d2a1(m-n)+d(m^2-m-n^2+n
假设m>nSn=A1+A2+……+AnSm=A1+A2+……+An+A(n+1)+A(n+2)+……+AmSm-Sn=A(n+1)+A(n+2)+……+Am=0(共m-n项)从A(n+1)项到Am项也
am=a1+(m-1)d=xan=a1+(n-1)d=y两式相减得(m-n)d=x-yd=(x-y)/(m-n)代入a1+(m-1)d=x得a1=(my-nx+x-y)/(m-n)所以ai=a1+(i
Am=A1+(m-1)d=pAn=A1+(n-1)d=qA(m+n)=A1+(m+n-1)dAm+An=2A1+(m+n-2)d=A1+(m+n-1)d+(A1-d)=p+qA(m+n)=A1+(m+
{log2(Xn)}为等差数列,故log2(Xn)-log2(Xn-1)=Xn/X(n-1)=q为常数故Xn为等比数列,由题可知:log2(Xm)=n,log2(Xn)=m,即Xm=2^n,Xn=2^