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Question-206992




Question Number 206992 by mnjuly1970 last updated on 03/May/24
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Answered by mr W last updated on 03/May/24
−1≤x^2 +x<0  ⇒−1<x<0  ⇒0<x^(2n) <1 ⇒⌊x^(2n) ⌋=0  ⇒−1<x^(2n+1) <0 ⇒⌊x^(2n+1) ⌋=−1  Σ_(k=1) ^(30) ⌊x^k ⌋=Σ_(n=1) ^(15) ⌊x^(2n) ⌋+Σ_(n=0) ^(14) ⌊x^(2n+1) ⌋                  =15×0+15×(−1)=−15
$$−\mathrm{1}\leqslant{x}^{\mathrm{2}} +{x}<\mathrm{0} \\ $$$$\Rightarrow−\mathrm{1}<{x}<\mathrm{0} \\ $$$$\Rightarrow\mathrm{0}<{x}^{\mathrm{2}{n}} <\mathrm{1}\:\Rightarrow\lfloor{x}^{\mathrm{2}{n}} \rfloor=\mathrm{0} \\ $$$$\Rightarrow−\mathrm{1}<{x}^{\mathrm{2}{n}+\mathrm{1}} <\mathrm{0}\:\Rightarrow\lfloor{x}^{\mathrm{2}{n}+\mathrm{1}} \rfloor=−\mathrm{1} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{30}} {\sum}}\lfloor{x}^{{k}} \rfloor=\underset{{n}=\mathrm{1}} {\overset{\mathrm{15}} {\sum}}\lfloor{x}^{\mathrm{2}{n}} \rfloor+\underset{{n}=\mathrm{0}} {\overset{\mathrm{14}} {\sum}}\lfloor{x}^{\mathrm{2}{n}+\mathrm{1}} \rfloor \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{15}×\mathrm{0}+\mathrm{15}×\left(−\mathrm{1}\right)=−\mathrm{15} \\ $$
Commented by mnjuly1970 last updated on 03/May/24
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$$\:\underline{\underbrace{\lesseqgtr}} \\ $$

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