设 S0=∫02f(x)e-xdx,S1=∫23f(x-2)e-xdx,…,Sn=∫2n2n+2f(x-2n)e-xdx,求Sn.

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问题
S0=∫02f(x)e-xdx,S1=∫23f(x-2)e-xdx,…,Sn=∫2n2n+2f(x-2n)e-xdx,求Sn

选项

答案S0=∫02f(x)e-xdx=∫01xe-xdx+∫12(2-x)e-xdx=(1-[*])2, 令t=x-2,则S1=e-202f(t)e-tdt=e-2S0, 令t=x-2n,则Sn=e-2n02f(t)e-tdt=e-2nS0, S=[*]Sn=S0[*]e-2n=[*]

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