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Find the sum to n terms of the series :`3xx8+6xx11+9xx14+dotdotdot` |
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Answer» Correct Answer - `3n(n+1)(n+3)` `T_(n)=("nth term of " 3,6,9,...)xx("nth term of "8,11, 14, ...)` `={3+(n-1)xx3}xx{8+(n-1)xx3}=3n(3n+5)=9n^(2)+15n.` `therefore S_(n) =9(sum_(k=1)^(n)k^(2))+15(sum_(k=1)^(n)k)`. |
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