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Sum the following series to `n`terms:`1/(1. 6)+1/(6. 11)+1/(11. 16)+1/(16. 21)++1/((5n-4)(5n+1))` |
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Answer» Correct Answer - `(n)/((5n+1))` `T_(n)=(1)/(5)xx{(1)/(5n-4)-(1)/(5n+1)}`. `therefore T_(1)=(1)/(5)xx{(1)/(1)-(1)/(6)}` `T_(2)=(1)/(5)xx{(1)/(6)-(1)/(11)}` `T_(3)=(1)/(5)xx{(1)/(11)-(1)/(16)}` ... ... ... ... ... ... ... ... ... ... `T_(n-1)=(1)/(5)xx{(1)/(5n-9)-(1)/(5n-4)}` `T_(n) =(1)/(5) xx {(1)/(5n-4)-(1)/(5n+1)}`. adding columnwise, we get `S_(n)=(T_(1)+T_(2)+ ... +T_(n))=(1)/(5)(1-(1)/(5n+1))=(5n+1-1)/(5(5n+1))=(n)/((5n+1))`. |
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