Find the sum of n terms of the series whose nth term is: `n(n+3)`
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We have , `t_(k)=k(k+3)=(k^(2)+3k)`.
`therefore ` sum to n terms is given by
`S_(n)=sum_(k=1)^(n)T_(k)`
`=sum_(k=1)^(n)(k^(2)+3k)=sum_(k=1)^(n)k^(2)+3sum_(k=1)^(n)k`
`=(1)/(6)n(n+1)(2n+1)+3*(1)/(2)n(n+1)=(1)/(6){n(n+1)(2n+1)+9n(n+1)}`
`=(1)/(6)n(n+1)(2n+1+9)=(1)/(6)n(n+1)*2(n+5)=(1)/(3)n(n+1)(n+5).`
Hence, the required sum is `(1)/(3)n(n+1)(n+5).`