f(x)=x2+x,则数列{1/fn}的前n项和为多少?我一年多没做过数列题了,
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![f(x)=x2+x,则数列{1/fn}的前n项和为多少?我一年多没做过数列题了,](/uploads/image/z/3951644-68-4.jpg?t=f%28x%29%3Dx2%2Bx%2C%E5%88%99%E6%95%B0%E5%88%97%7B1%2Ffn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BA%E5%A4%9A%E5%B0%91%3F%E6%88%91%E4%B8%80%E5%B9%B4%E5%A4%9A%E6%B2%A1%E5%81%9A%E8%BF%87%E6%95%B0%E5%88%97%E9%A2%98%E4%BA%86%2C)
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f(x)=x2+x,则数列{1/fn}的前n项和为多少?我一年多没做过数列题了,
f(x)=x2+x,则数列{1/fn}的前n项和为多少?
我一年多没做过数列题了,
f(x)=x2+x,则数列{1/fn}的前n项和为多少?我一年多没做过数列题了,
利用差分法
1/fn=1/(n²+n)=1/(n*(n+1))=1n-1/(n+1)
所以前n项和
Sn=(1-1/2)+(1/2-1/3)+...+(1/n-1/(n+1))
=1-1/(n+1)=n/(n+1)
f(n)=n²+n=n*(n+1)
1/f(n)=1/[n*(n+1)]=1/n-1/(n+1)
Sn=f(1)+f(2)+f(3)+······f(n)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+······[1/n-1/(n+1)]
=1-1/2+1/2-1/3+1/3-1/4+······1/n-1/(n+1)
=1-1/(n+1)=n/(n+1)