1个回答
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1+x+x^2+..+x^n = [x^(n+1) -1]/(x-1)
put x =e^(1/n)
1+e^(1/n) +..+e^(1) = [(e^(1/n))^(n+1) -1]/(e^(1/n)-1)
put x =e^(1/n)
1+e^(1/n) +..+e^(1) = [(e^(1/n))^(n+1) -1]/(e^(1/n)-1)
追问
i=1,从1开始取值,为什么不是e^1/n+...+e^(1) .为什么前面要算上1
追答
divided [0,1] into n equal interval with width 1/n
∫(0->1) e^x dx
=lim(n->无穷)summation(i:1->n) (1/n) e^(i/n)
=lim(n->无穷) [ 1/n [ e^(1/n) + e^(2/n)+.... +e^(n/n) ]
consider
1+x+x^2+..+x^n = [x^(n+1) -1]/(x-1)
1+x+x^2+..+x^n = [x^(n+1) -1]/(x-1)
put x =e^(1/n)
1+e^(1/n) +..+e^(1)= [(e^(1/n))^(n+1) -1]/(e^(1/n)-1)
e^(1/n) +..+e^(1) = [(e^(1/n))^(n+1) -1]/(e^(1/n)-1) -1
∫(0->1) e^x dx
=lim(n->无穷) [ 1/n [ e^(1/n) + e^(2/n)+.... +e^(n/n) ]
=lim(n->无穷) [ 1/n [ [(e^(1/n))^(n+1) -1]/(e^(1/n)-1) -1 ]
=lim(n->无穷) [ 1/n [ [(e^(1/n))^(n+1) -1]/(e^(1/n)-1) ] - lim(n->无穷)(1/n)
= e-1
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