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Find the sum to n terms of the series , ...

Find the sum to n terms of the series , 5+11+19+29+41…

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Let us write `S_(n)=5+11+19+29+....+a_(n-1)+a_(n)`
or `S_(n)=5+11+19+.....a_(n-2)+a_(n-1)+a_(n)`
On subtration , we get
`0=5+[6+8+10+12+....(n-1)"terms"]-a_(n)`
or `a_(n)=5+((n-1)[12+(n-2)xx2])/(2)`
`=5+(n-1)(n+4)=n^(2)+3n+1`
Hence `S_(n)=sum_(k=1)^(n)a_(x)sum_(k=1)^(n)(k^(2)+3k+1)=sum_(k=1)^(n)k^(2)+3sum_(1)^(n)k+n`
`=(n(n+1)(2n+1))/(6)+(3n(n+1))/(2)+n=(n(n+2)(n+4))/(3)`
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