数列求和,有数列A(n),A(1)=i,A(2)=A(1)+r,A(3)=A(2)+2r,……,A(n)=A(n-1)+(n-1)*r,求前N项和公式如果i=5,r=1,数列是这样:5,6,8,11,15,20,26,33,41,50,60,71,83,96,...如果i=10,r=5,数列是这样:10,15,25,40,60,85,115,

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数列求和,有数列A(n),A(1)=i,A(2)=A(1)+r,A(3)=A(2)+2r,……,A(n)=A(n-1)+(n-1)*r,求前N项和公式如果i=5,r=1,数列是这样:5,6,8,11,15,20,26,33,41,50,60,71,83,96,...如果i=10,r=5,数列是这样:10,15,25,40,60,85,115,

数列求和,有数列A(n),A(1)=i,A(2)=A(1)+r,A(3)=A(2)+2r,……,A(n)=A(n-1)+(n-1)*r,求前N项和公式如果i=5,r=1,数列是这样:5,6,8,11,15,20,26,33,41,50,60,71,83,96,...如果i=10,r=5,数列是这样:10,15,25,40,60,85,115,
数列求和,
有数列A(n),A(1)=i,A(2)=A(1)+r,A(3)=A(2)+2r,……,A(n)=A(n-1)+(n-1)*r,求前N项和公式
如果i=5,r=1,数列是这样:5,6,8,11,15,20,26,33,41,50,60,71,83,96,...
如果i=10,r=5,数列是这样:10,15,25,40,60,85,115,150,190,235,...
如果i=1,r=5,数列是这样:1,6,16,31,51,76,106,141,181,...
@wgl5411:怎么会是等差数列呢?你好好看清楚
@赤之墨:你这是S(n)通项公式,还是前n项之和?

数列求和,有数列A(n),A(1)=i,A(2)=A(1)+r,A(3)=A(2)+2r,……,A(n)=A(n-1)+(n-1)*r,求前N项和公式如果i=5,r=1,数列是这样:5,6,8,11,15,20,26,33,41,50,60,71,83,96,...如果i=10,r=5,数列是这样:10,15,25,40,60,85,115,
A(n)-A(n-1)=(n-1)*r;
A(n-1)-A(n-2)=(n-2)*r,
.
A(2)-A(1)=r;
全部相加得,A(n)-A(1)=n*(n-1)*r/2;
化简得A(n)=i+n(n-1)r/2=i+nr/2+n^2*r/2
所以前N项和Tn=ni+n(n+1)r/4+n(n-1)(2n-1)r/12

S(n)=S(n-1)+a+(n-1)*r
所以S(n)-S(n-1)=a+(n-1)*r
即a(n)=a+(n-1)*r
说明数列是等差数列
由等差树列求和公式得
Sn=na+n(n-1)r/2

s(n)-s(n-1)=a+(n-1)*r;s(n-1)-s(n-2)=a+(n-2)*r,........s(2)-s(1)=a+r;累加得,s(n)-s(1)=a*(n-1)+(n-1)*(n-2)*r/2;化简得s(n)=a*n+(n-1)*(n-2)*r/2

sfd

全部展开,A(n)=an^4+bn^3+cn^2+dn+6然后分4个数列求和,前面系数提出来就是单阶的求和了,都有公式吧