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3!+4!2(1!+2!)+4!+5!3(2!+3!)+…+102!+103!101(100!+101!)= ?\frac{3!+4!}{2(1!+2!)}+\frac{4!+5!}{3(2!+3!)}+\ldots+\frac{102!+103!}{101(100!+101!)} = \ ? 2(1!+2!)3!+4!+3(2!+3!)4!+5!+…+101(100!+101!)102!+103!= ?
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