an的前n项和sn=(3/2)n^2+59/2n 求|a1|+|a2|+|a3|+...+|an|

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an的前n项和sn=(3/2)n^2+59/2n 求|a1|+|a2|+|a3|+...+|an|
an的前n项和sn=(3/2)n^2+59/2n 求|a1|+|a2|+|a3|+...+|an|

an的前n项和sn=(3/2)n^2+59/2n 求|a1|+|a2|+|a3|+...+|an|
sn=(3/2)n^2+59/2n
an=sn-s(n-1)=3n+28
所以an>0
所以原式=a1+a2+……+an=sn=(3/2)n^2+59/2n

sinx+siny=√2 (1)
cosx+cosy=2√3/3 (2)
(1)^2+(2)^2
2+2sinxsiny+2cosxcosy=2+2/3
cosxcosy+sinxsiny=2/3 (3)
cos (x-y)=2/3 (4)
(2)^2-(1)^2
cos2x+cos2y+2(cosxcosy-sinxsiny)=-2...

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sinx+siny=√2 (1)
cosx+cosy=2√3/3 (2)
(1)^2+(2)^2
2+2sinxsiny+2cosxcosy=2+2/3
cosxcosy+sinxsiny=2/3 (3)
cos (x-y)=2/3 (4)
(2)^2-(1)^2
cos2x+cos2y+2(cosxcosy-sinxsiny)=-2/3
2cos(x+y)cos(x-y)+2cos(x+y)=-2/3
把(4)代入,得到cos(x+y)=-1/5
所以cosxcosy-sinxsiny=-1/5 (5)
由(3)和(5)解得cosxcosy=7/30,sinxsiny=13/30
所以tanxtany=sinxsiny/cosxcosy=13/7

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