数学数列累加求和已知n!=1X2X3X4X.Xn,求1X1!+2X2!+3X3!+4X4!+.+nXn!=?

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数学数列累加求和已知n!=1X2X3X4X.Xn,求1X1!+2X2!+3X3!+4X4!+.+nXn!=?
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数学数列累加求和已知n!=1X2X3X4X.Xn,求1X1!+2X2!+3X3!+4X4!+.+nXn!=?
数学数列累加求和
已知n!=1X2X3X4X.Xn,求1X1!+2X2!+3X3!+4X4!+.+nXn!=?

数学数列累加求和已知n!=1X2X3X4X.Xn,求1X1!+2X2!+3X3!+4X4!+.+nXn!=?
做求和的题,首先就化简通向公司,nXn!=(n+1-1)n!=(n+1)!-n!然后就不用说了吧

也就等于;

n╳(n+1)╳n!╳1/2

(1*1!+2*2!+3*3!+4*4!+.......+n*n!)+(1!+2!+3!+4!+.......+n!)
=2!+3!+4!+.......+n!+(n+1)!
所以:
1*1!+2*2!+3*3!+4*4!+.......+n*n!
=[2!+3!+4!+.......+n!+(n+1)!]-[1!+2!+3!+4!+.......+n!]
=(n+1)!-1