计算一道数学题 用简便算法 1/(1×2×3 )+ 1(2×3×4) +…+ 1/(98×99×100)
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计算一道数学题 用简便算法 1/(1×2×3 )+ 1(2×3×4) +…+ 1/(98×99×100)
计算一道数学题 用简便算法 1/(1×2×3 )+ 1(2×3×4) +…+ 1/(98×99×100)
计算一道数学题 用简便算法 1/(1×2×3 )+ 1(2×3×4) +…+ 1/(98×99×100)
首先写出这个式子的通项
a(n)=1/(n*(n+1)*(n+2))
所以
a(n)+a(n+1)
=1/(n*(n+1)*(n+2))+1/((n+1)*(n+2)*(n+3))
=(2n+3)/(n*(n+1)*(n+2)*(n+3))
=1/(n*(n+2))-1/((n+1)*(n+3))
写成这个样子
就很简单了
a1+a2=1/1*3-1/2*4
a2+a3=1/2*4-1/3*5
……
所以
(a1+a2)+(a2+a3)+…+(a97+a98)
=(1/1*3-1/2*4)+(1/2*4-1/3*5)+…+(1/97*99-1/98*100)
=1/1*3-1/98*100
这个结果加上头尾两个
即a1和a98就是题目所求的两倍
(1/3-1/9800+1/6+1/9800*99)/2
=(1/2-1/9900)/2
=4959/19800
答案是 4959/19800
a(n)=1/(n*(n+1)*(n+2))
所以
a(n)+a(n+1)
=1/(n*(n+1)*(n+2))+1/((n+1)*(n+2)*(n+3))
=(2n+3)/(n*(n+1)*(n+2)*(n+3))
=1/(n*(n+2))-1/((n+1)*(n+3))
a1+a2=1/1*3-1/2*4
a2+a3=1/2...
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a(n)=1/(n*(n+1)*(n+2))
所以
a(n)+a(n+1)
=1/(n*(n+1)*(n+2))+1/((n+1)*(n+2)*(n+3))
=(2n+3)/(n*(n+1)*(n+2)*(n+3))
=1/(n*(n+2))-1/((n+1)*(n+3))
a1+a2=1/1*3-1/2*4
a2+a3=1/2*4-1/3*5
……
所以
(a1+a2)+(a2+a3)+…+(a97+a98)
=(1/1*3-1/2*4)+(1/2*4-1/3*5)+…+(1/97*99-1/98*100)
=1/1*3-1/98*100
(1/3-1/9800+1/6+1/9800*99)/2
=(1/2-1/9900)/2
=4959/19800
答案是 4959/19800
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