数学题1/(1+2)+1/(1+2+3)+····+1/(1+2+3+n)=?
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数学题1/(1+2)+1/(1+2+3)+····+1/(1+2+3+n)=?
数学题1/(1+2)+1/(1+2+3)+····+1/(1+2+3+n)=?
数学题1/(1+2)+1/(1+2+3)+····+1/(1+2+3+n)=?
1/(1+2+3.+n)=1/(n+1)n/2=2[1/n-1/(n+1)]
原式=2[1/2-1/3+1/3-1/4.+1/n-1/(n+1)]=2[1/2-1/(n+1)]=(n-1)/(n+1)
1/(1+2)+1/(1+2+3)+····+1/(1+2+3+n)
=2/2*3+2/3*4+……+2/n(n+1)
=2[1/2-1/3+1/3-1/4+……+1/n-1/(n+1)]
=2[1/2-1/(n+1)]
=(n-1)/(n+1)