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Question

Find the sum of the series 1·n+2(n-1)+3(n-2)+...+(n-1)2+n·1


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Solution

Step 1: Given

Let rth term be Tr

Tr=r[n-(r-1)]=rn-r2-r

Step 2: Sum Tr by sigma method

Let Sn be the sum of the series up to n terms

Sn=r=1nTrSn=r=1n(rn-r2-r)Sn=nr-r=1nr2r=1n-rr=1nSn=nn(n+1)2-n(n+1)(2n+1)6-n(n+1)2k=1nk=kk+12,k=1nk2=kk+12k+16Sn=nn+12n-2n+13-1Sn=nn+123n-2n-1-33Sn=nn+1n-46

Therefore, 1·n+2(n-1)+3(n-2)+...+(n-1)2+n·1=nn+1n-46


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