Sn=1*2*3+2*3*4+3*4*5+...+n(n+1)(n+2)求Sn
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Sn=1*2*3+2*3*4+3*4*5+...+n(n+1)(n+2)求Sn
![Sn=1*2*3+2*3*4+3*4*5+...+n(n+1)(n+2)求Sn](/uploads/image/z/17984286-54-6.jpg?t=Sn%3D1%2A2%2A3%2B2%2A3%2A4%2B3%2A4%2A5%2B...%2Bn%28n%2B1%29%28n%2B2%29%E6%B1%82Sn)
n*(n+1)*(n+2)=1/4*n*(n+1)*(n+2)[n+3-(n-1)]
Sn=1*2*3+2*3*4+3*4*5+...+n*(n+1)*(n+2)
=1/4{1*2*3*(4-0)+2*3*4*(5-1)+3*4*5*(6-2)...+n*(n+1)*(n+2)[n+3-(n-1)]}
如此裂项相消
原式= n*(n+1)*(n+2)*(n+3)/4
Sn=1*2*3+2*3*4+3*4*5+...+n*(n+1)*(n+2)
=1/4{1*2*3*(4-0)+2*3*4*(5-1)+3*4*5*(6-2)...+n*(n+1)*(n+2)[n+3-(n-1)]}
如此裂项相消
原式= n*(n+1)*(n+2)*(n+3)/4
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