Question

高数题！图片中的16 20 21题如何解答 求数学高手！！急急急！！

Answer 1

Answer 2

16.
1/[(n+1)(n+2)]+1/[(n+2)(n+3)]+...+1/[(n+n)(n+n+1)]<1/(n+1)²+ 1/(n+2)²+...+1/(n+n)²<1/[n(n+1)]+1/[(n+1)(n+2)]+...+1/[(n+n-1)(n+n)]
1/(n+1)-1/(n+2)+1/(n+2)-1/(n+3)+...+1/(n+n)-1/(n+n+1)<1/(n+1)²+ 1/(n+2)²+...+1/(n+n)²<1/n-1/(n+1)+1/(n+1)-1/(n+2)+...+1/(n+n-1)-1/(n+n)
1/(n+1)- 1/(2n+1)<1/(n+1)²+ 1/(n+2)²+...+1/(n+n)²<1/n- 1/(2n)
n/[(n+1)(2n+1)]<1/(n+1)²+ 1/(n+2)²+...+1/(n+n)²<1/(2n)
lim n/[(n+1)(2n+1)]
n→∞
=lim 1/[(1+ 1/n)(2n+1)]
n→∞
=lim 1/(2n+1)
n→∞
=0
lim1/(2n)=0
n→∞
由夹逼准则，得：
lim [1/(n+1)²+ 1/(n+2)²+...+1/(n+n)²]=0
n→∞
20.
∫[0：1]e^(πx)cos(πx)dx
=(1/π)∫[0：1]e^(πx)cos(πx)d(πx)
令πx=u
x：0→1，u：0→π
∫[0：1]e^(πx)cos(πx)dx
=(1/π)∫[0：π]e^ucosudu
=(1/π)∫[0：π]e^ud(sinu)
=(1/π)(e^usinu)|[0：π]-(1/π)∫[0：π]sinud(e^u)
=(1/π)(e^π·sinπ-e⁰·sin0) +(1/π)∫[0：π]e^u(-sinu)du
=0+(1/π)∫[0：π]e^ud(cosu)
=(1/π)e^ucosu|[0：π]-(1/π)∫[0：π]cosud(e^u)
=(1/π)(e^πcosπ-e⁰cos0)-(1/π)∫[0：π]e^ucosudu
=(1/π)(-e^π-1)-(1/π)∫[0：π]e^ucosudu
(2/π)∫[0：π]e^ucosudu=(1/π)(-e^π-1)
∫[0：π]e^ucosudu=-½(1+e^π)
∫[0：1]e^(πx)cos(πx)dx=-½(1+e^π)
21.
dy/dx=y/√(1-x²)
(1/y)dy=[1/√(1-x²)]dx
两边同时积分，得：
lny=arcsinx +C
y=e^(arcsinx +C)

追问

多谢！！！

追答

别客气。你选择的这个人回答得很好。步骤比我的还要简便些。