1.

Sum the series `3. 8+6. 11+9. 14+ `to `n`terms.

Answer» We have
`T_(k)=("kth term of "3,6,9, …) xx ("kth term of "8,11,14,…)`
`={3+(k-1)xx3}xx[8+(k-1)xx 3} = 3k(3k+5)`
`=(9k^(2)+15k).`
`S_(n)=sum_(k=1)^(n)T_(k)`
`=sum_(k=1)^(n)(9k^(2)+15k)=9(sum_(k=1)^(n)k^(2))+15(sum_(k=1)^(n)k)`
`=9*{(1)/(6)n(n+1)(2n+1)}+15*{(1)/(2)n(n+1)}`
`=(3)/(2)n(n+1){(2n+1)+5}=3n(n+1)(n+3)`.
Hence, the required sum is `3n(n+1)(n+3)`.


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