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Question Number 202306 by hardmath last updated on 24/Dec/23
Find the 2023rd term in the sequence 2,3,5,6,7,8,10,11,12,13,14,15,17,18,... obtained by subtracting integer squares from natural numbers.
$$ \\ $$Find the 2023rd term in the sequence 2,3,5,6,7,8,10,11,12,13,14,15,17,18,… obtained by subtracting integer squares from natural numbers.
Answered by AST last updated on 24/Dec/23
1,2,3,...,2023 contains 44 squares ⇒2023 is the (2023−44=1979)th term 2023−1979th term 2023+44−2022nd term; since 45^2 =2025 ⇒2023rd term=2023+45=2068
$$\mathrm{1},\mathrm{2},\mathrm{3},…,\mathrm{2023}\:{contains}\:\mathrm{44}\:{squares} \\ $$$$\Rightarrow\mathrm{2023}\:{is}\:{the}\:\left(\mathrm{2023}−\mathrm{44}=\mathrm{1979}\right){th}\:{term} \\ $$$$\mathrm{2023}−\mathrm{1979}{th}\:{term} \\ $$$$\mathrm{2023}+\mathrm{44}−\mathrm{2022}{nd}\:{term};\:{since}\:\mathrm{45}^{\mathrm{2}} =\mathrm{2025} \\ $$$$\Rightarrow\mathrm{2023}{rd}\:{term}=\mathrm{2023}+\mathrm{45}=\mathrm{2068} \\ $$
Commented by hardmath last updated on 24/Dec/23
thank you dear professor
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$
Answered by mr W last updated on 24/Dec/23
1,2,3,4,5,...,n 1^2 ,2^2 ,3^2 ,...,k^2 with k^2 <n, i.e. k=⌊(√n)⌋ m=n−k=n−⌊(√n)⌋ that means the m^(th) term is n. ⌊(√n)⌋≤(√n) m=n−⌊(√n)⌋≥n−(√n) n−(√n)−m≤0 ⇒(√n)≤((1+(√(1+4m)))/2) ⇒n≤((((√(4m+1))+1)^2 )/4) i.e. the m^(th) term is a_m =⌊((((√(4m+1))+1)^2 )/4)⌋ examples: m=2023: a_(2023) =⌊((((√(4×2023+1))+1)^2 )/4)⌋=2068 m=2025: a_(2025) =⌊((((√(4×2025+1))+1)^2 )/4)⌋=2070 m=1000: a_(1000) =⌊((((√(4×1000+1))+1)^2 )/4)⌋=1032 m=3000: a_(3000) =⌊((((√(4×3000+1))+1)^2 )/4)⌋=3055
$$\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},…,{n} \\ $$$$\mathrm{1}^{\mathrm{2}} ,\mathrm{2}^{\mathrm{2}} ,\mathrm{3}^{\mathrm{2}} ,…,{k}^{\mathrm{2}} \:{with}\:{k}^{\mathrm{2}} <{n},\:{i}.{e}.\:{k}=\lfloor\sqrt{{n}}\rfloor \\ $$$${m}={n}−{k}={n}−\lfloor\sqrt{{n}}\rfloor \\ $$$${that}\:{means}\:{the}\:{m}^{{th}} \:{term}\:{is}\:{n}. \\ $$$$\lfloor\sqrt{{n}}\rfloor\leqslant\sqrt{{n}} \\ $$$${m}={n}−\lfloor\sqrt{{n}}\rfloor\geqslant{n}−\sqrt{{n}} \\ $$$${n}−\sqrt{{n}}−{m}\leqslant\mathrm{0} \\ $$$$\Rightarrow\sqrt{{n}}\leqslant\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4}{m}}}{\mathrm{2}}\: \\ $$$$\Rightarrow{n}\leqslant\frac{\left(\sqrt{\mathrm{4}{m}+\mathrm{1}}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$${i}.{e}.\:{the}\:{m}^{{th}} \:{term}\:{is} \\ $$$${a}_{{m}} =\lfloor\frac{\left(\sqrt{\mathrm{4}{m}+\mathrm{1}}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}\rfloor \\ $$$${examples}: \\ $$$${m}=\mathrm{2023}:\: \\ $$$$\:\:\:{a}_{\mathrm{2023}} =\lfloor\frac{\left(\sqrt{\mathrm{4}×\mathrm{2023}+\mathrm{1}}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}\rfloor=\mathrm{2068} \\ $$$${m}=\mathrm{2025}:\: \\ $$$$\:\:\:{a}_{\mathrm{2025}} =\lfloor\frac{\left(\sqrt{\mathrm{4}×\mathrm{2025}+\mathrm{1}}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}\rfloor=\mathrm{2070} \\ $$$${m}=\mathrm{1000}: \\ $$$$\:\:\:{a}_{\mathrm{1000}} =\lfloor\frac{\left(\sqrt{\mathrm{4}×\mathrm{1000}+\mathrm{1}}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}\rfloor=\mathrm{1032} \\ $$$${m}=\mathrm{3000}: \\ $$$$\:\:\:{a}_{\mathrm{3000}} =\lfloor\frac{\left(\sqrt{\mathrm{4}×\mathrm{3000}+\mathrm{1}}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}\rfloor=\mathrm{3055} \\ $$
Commented by hardmath last updated on 24/Dec/23
perfect solution tahnkyou dear professor
$$\mathrm{perfect}\:\mathrm{solution}\:\mathrm{tahnkyou}\:\mathrm{dear}\:\mathrm{professor} \\ $$

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