lim(cosx+cos2x+ cos3x+…+ cos nx-n)/(cosx-1) (x趋于0)
lim(cosx+cos2x+ cos3x+…+ cos nx-n)/(cosx-1) (x趋于0)
2,3,n之类的是幂,不是2乘以x
2,3,n之类的是幂,不是2乘以x
赞 BBTwawa 幼苗
共回答了21个问题采纳率：90.5% 举报
consider
1+2+3+..+n = n(n+1)/2
n^2 = n(n+1)-n
= (1/3) [n(n+1)(n+2)-(n-1)n(n+1)] -n
1^2+2^2+3^2+..+n^2
=(1/3)n(n+1)(n+2) -n(n+1)/2
=(1/6)n(n+1)(2n+1)
lim(x->0) (cosx+cos2x+ cos3x+…+ cosnx-n)/(cosx-1) (0/0)
= lim(x->0) (sinx+2sin2x+ 3sin3x+…+ nsinnx)/(sinx) (0/0)
=lim(x->0) (cosx+2^2cos2x+ 3^3cos3x+…+ n^2cosnx)/(cosx)
=1^2+2^2+..+n^2
=(1/6)n(n+1)(2n+1)
1+2+3+..+n = n(n+1)/2
n^2 = n(n+1)-n
= (1/3) [n(n+1)(n+2)-(n-1)n(n+1)] -n
1^2+2^2+3^2+..+n^2
=(1/3)n(n+1)(n+2) -n(n+1)/2
=(1/6)n(n+1)(2n+1)
lim(x->0) (cosx+cos2x+ cos3x+…+ cosnx-n)/(cosx-1) (0/0)
= lim(x->0) (sinx+2sin2x+ 3sin3x+…+ nsinnx)/(sinx) (0/0)
=lim(x->0) (cosx+2^2cos2x+ 3^3cos3x+…+ n^2cosnx)/(cosx)
=1^2+2^2+..+n^2
=(1/6)n(n+1)(2n+1)
1年前
8
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你能帮帮他们吗