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Find the sum to n terms of the series 1x...

Find the sum to n terms of the series `1xx2+2xx3+ 3xx4+ 4xx5+......`

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The correct Answer is:
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Let `S_(n)=1xx2+2xx3+(3xx4)+(4xx5)+…+` to n terms.
nth term of given sequence `=n(n+1)=n^(2)+n`
`:. S_(n)=sum(n^(2)+n)=sumn^(2)+sumn`
`=(n(n+1)(2n+1))/(6)+(n(n+1))/(2)`
`=(n(n+1))/(2)[(2n+1)/(3)+1]=(n(n+1))/(2)xx(2n+4)/(3)`
`=(n(n+1)(n+2))/(3)`
Therefore , the sum of n terms of given sequence `=(n(n+1)(n+2))/(3)`
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