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

Find the sum of the series `1*n+2*(n-1)+3*(n-2)+4*(n-3)+....(n−1).2+n.1"`

Text Solution

Verified by Experts

The rth term of the given series is
`T_(r)=r*(n-r+1)=(n+1)r-r^(2)`
`therefore` Sum of the series
`S_(n)=sum _(r=1)^(n)T_(r)=(n+1)sum _(r=1)^(n)r-sum _(r=1)^(n)r^(2)=(n+1)sumn-sumn^(2)`
`=(n+1)(n(n+1))/(2)-(n(n+1)(2n+1))/(6)`
`=(n(n+1))/(6)(3n+3-2n-1)=(n(n+1)(n+2))/(6)`
Now,
`(1+2x+3x^(2)+"..."+nx^(n-1))^(2)=(1+2x+3x^(2)+"..."+nx^(n-1))xx(1+2x+3x^(2)+"..."+nx^(n-1))`
`therefore " Coeffocient of "x^(n-1) " in "(1+2x+3x^(2)+"..."+nx^(n-1))^(2)`
`1*n+2*(n-1)+3*(n-2)+"..."+n*1`
`S_(n)=(n(n+1)(n+2))/(6)`
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