如图,求数列{1/64,1/32,1/16,}的前n项和。
2个回答
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1.1/2+1/4+1/8+1/16+1/32+1/64
=32/64+16/64+8/64+4/64+2/64+1/64
=(32+16+8+4+2+1)/64
=63/64
2.1/2+1/4+1/8+1/16+1/32+1/64
=(1-1/2)+(1/2-1/4)+(1/4-1/8)+(1/8-1/16)+(1/16-1/32)+(1/32-1/64)
=1-1/2+1/2-1/4+1/4-1/8+1/8-1/16+1/16-1/32+1/32-1/64
=1-1/64
=63/64
3.原式=1/2+1/4+1/8+1/16+1/32+1/64+1/64-1/64=1-1/64=63/64
4原式
=(32+16+8+4+2+1)/64
=63/64;
5、令m=1/2+1/4+1/8+1/16+1/32+1/64
两边同时乘2,得:
2m=1+1/2+1/4+1/8+...+1/32
两式相减,得:
m=1-1/64=63/646、作一矩形,(对不起,发不了图)对折得1/2,再对折得1/4,再对折得1/8,再对折得1/16,再对折得1/32,再对折得1/64,和为63/64
7、
这个是一个公比为1/2的等比数列求和
Sn=(a1-an×q)/(1-q)=(1/2-1/64×1/2)/(1-1/2)=63/ 64
8、原式=(1/2+1/4)+1/8+1/16+1/32+1/64
=(3/4+1/8)+1/16+1/32+1/64
=(7/8+1/16)+1/32+1/64
=. . .
=63/64
=32/64+16/64+8/64+4/64+2/64+1/64
=(32+16+8+4+2+1)/64
=63/64
2.1/2+1/4+1/8+1/16+1/32+1/64
=(1-1/2)+(1/2-1/4)+(1/4-1/8)+(1/8-1/16)+(1/16-1/32)+(1/32-1/64)
=1-1/2+1/2-1/4+1/4-1/8+1/8-1/16+1/16-1/32+1/32-1/64
=1-1/64
=63/64
3.原式=1/2+1/4+1/8+1/16+1/32+1/64+1/64-1/64=1-1/64=63/64
4原式
=(32+16+8+4+2+1)/64
=63/64;
5、令m=1/2+1/4+1/8+1/16+1/32+1/64
两边同时乘2,得:
2m=1+1/2+1/4+1/8+...+1/32
两式相减,得:
m=1-1/64=63/646、作一矩形,(对不起,发不了图)对折得1/2,再对折得1/4,再对折得1/8,再对折得1/16,再对折得1/32,再对折得1/64,和为63/64
7、
这个是一个公比为1/2的等比数列求和
Sn=(a1-an×q)/(1-q)=(1/2-1/64×1/2)/(1-1/2)=63/ 64
8、原式=(1/2+1/4)+1/8+1/16+1/32+1/64
=(3/4+1/8)+1/16+1/32+1/64
=(7/8+1/16)+1/32+1/64
=. . .
=63/64
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