Calculate 1 + (1 + 2) + (1 + 2 + 3) + +(1+2+3+… +n).
∵1+2+3+… +n=n(n+1)2=n2+n2,∴1+(1+2)+(1+2+3)+… +(1+2+3+… +n)=12(1+12+2+22+3+32+… +n+n2)=12[(1+2+3+… +n)+(12+22+32+… +n2)]=12•[n(n+1)2+n(n+1)(2n+1)6]=n(n+1)4+n(n+1)(2n+1)12....
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