计算 (1+1/1*3)*(1+1/2*4)*(1+1/3*5).*(1+1/98*100)
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计算 (1+1/1*3)*(1+1/2*4)*(1+1/3*5).*(1+1/98*100)
计算 (1+1/1*3)*(1+1/2*4)*(1+1/3*5).*(1+1/98*100)
计算 (1+1/1*3)*(1+1/2*4)*(1+1/3*5).*(1+1/98*100)
1+1/[n(n+2)] = (n+1)^2 / [n(n+2)]
所以原式等于2*99/100
(1+1/1×3)×(1+1/2×4)×(1+1/3×5)×(1+1/4×6)×……×(1+1/97×99)×(1+1/98×100)
=(1×1+1×1/3)×(2×1/2+1/2×1/4)×(3×1/3+1/3×1/5)×(4×1/4+1/4×1/6)×……×(97×1/97+1/97×1/99)×(98×1/98+1/98×1/100)
=[1×(1+1/3)]×[1/2×(...
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(1+1/1×3)×(1+1/2×4)×(1+1/3×5)×(1+1/4×6)×……×(1+1/97×99)×(1+1/98×100)
=(1×1+1×1/3)×(2×1/2+1/2×1/4)×(3×1/3+1/3×1/5)×(4×1/4+1/4×1/6)×……×(97×1/97+1/97×1/99)×(98×1/98+1/98×1/100)
=[1×(1+1/3)]×[1/2×(2+1/4)]×[1/3×(3+1/5)]×[1/4×(4+1/6)]×……×[1/97×(97+1/99)]×[1/98×(98+1/100)]
=(2²/3×3²/4×4²/5×5²/6×……×98²/99×99²/100)/(1×2×3×4×……×97×98)
=(2²×3²×4²×5²×……×98²×99²)/(1×2×3²×4²×……×97²×98²×99×100)
=(2×99)/(1×100)
=198/100
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显然通项可以表示为：1+1/n(n+2)=[n(n+2)+1]/n(n+2)=(n^2+2n+1)/n(n+2)=(n+1)/n*(n+1)/(n+2)
所以原式=[2/1*2/3]*[3/2*3/4]*[4/3*4/5]*......*[98/97*98/99]*[99/98*99/100]
=2/1*99/100
=99/50