1/2×[1+2+3...+(n+1)+1+2+3...+(n+1)]=1/2(n+1)(n+2)的详细解答过程
好吧,你采纳我。
原式=1/2×[2×(1+2+3+4+5+······+(n+1))]①
=1+2+3+4+5+······+(n+1)②
=1/2×[(n+1+1)×(n+1)]③
=1/2(n+1)(n+2)④
在②到③时用到了高斯算法,就是首相加末相乘以项数除以二。可以百度的。
不懂请追问。望采纳。谢谢。
原式=1/2[(1+n+1)×(n+1)/2+(1+n+1)×)×(n+1)/2]
=1/2[(n+2)(n+1)/2+(n+2)(n+1)/2]
=1/2(n+2)(n+1)
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