设bn=b1+d(b1 、 d 均为字母表示的已知条件),又设an=cosbn ,求an前n项和Sn表达式
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设bn=b1+d(b1 、 d 均为字母表示的已知条件),又设an=cosbn ,求an前n项和Sn表达式
bn=b1+(n-1)*d 打错了
bn=b1+(n-1)*d 打错了
利用到下面二个公式
sinx+sin2x+sin3x+……+sinnx=[sin(nx/2)sin((n+1)x/2)]/sin(x/2)
cosx+cos2x+cos3x+……+cosnx=[cos((n+1)x/2)sin(nx/2)]/sin(x/2)
an=cosbn
a1=cosb1
a2=cos(b1+d)
a3=cos(b1+2d)
.
an=cos(b1+(n-1)d)
Sn=a1+a2+.+an
=cosb1+cos(b1+d)+cos(b1+2d)+.+cos(b1+(n-1)d)
=cosb1+cosb1cosd-sinb1sind +cosb1cos2d-sinb1sin2d+.+cosb1cos(n-1)d -sinb1sin(n-1)d
=cosb1(1+cosd+cos2d+.+cos(n-1)d)-sinb1(sind+sin2d+sin3d+..+sin(n-1)d)
=cosb1(1+[sin((n-1)d/2) sin(nd/2)]/sin(d/2) -sinb1(cos((nd/2) sin((n-1)d/2)]/sin(d/2)
sinx+sin2x+sin3x+……+sinnx=[sin(nx/2)sin((n+1)x/2)]/sin(x/2)
cosx+cos2x+cos3x+……+cosnx=[cos((n+1)x/2)sin(nx/2)]/sin(x/2)
an=cosbn
a1=cosb1
a2=cos(b1+d)
a3=cos(b1+2d)
.
an=cos(b1+(n-1)d)
Sn=a1+a2+.+an
=cosb1+cos(b1+d)+cos(b1+2d)+.+cos(b1+(n-1)d)
=cosb1+cosb1cosd-sinb1sind +cosb1cos2d-sinb1sin2d+.+cosb1cos(n-1)d -sinb1sin(n-1)d
=cosb1(1+cosd+cos2d+.+cos(n-1)d)-sinb1(sind+sin2d+sin3d+..+sin(n-1)d)
=cosb1(1+[sin((n-1)d/2) sin(nd/2)]/sin(d/2) -sinb1(cos((nd/2) sin((n-1)d/2)]/sin(d/2)
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