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