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Question Number 206038 by BaliramKumar last updated on 05/Apr/24

$$\sqrt{1+2023\sqrt{1+2024\sqrt{1+2025\sqrt{1+2026\sqrt{1+..............\mathrm{\infty }}}}}}=?$$

Answered by mr W last updated on 05/Apr/24

$$x+1=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+...}}}}$$$$\Rightarrow 2024=\sqrt{1+2023\sqrt{1+2024\sqrt{1+2025\sqrt{1+2026\sqrt{1+..............\mathrm{\infty }}}}}}$$

Answered by MATHEMATICSAM last updated on 05/Apr/24

$$\mathbf{This}\mathbf{problem}\mathbf{is}\mathbf{based}\mathbf{on}{\mathbf{Ramanujan}}^{\prime }\mathbf{s}$$$$\mathbf{infinite}\mathbf{nested}\mathbf{redical}\mathbf{problem}.$$$${(n+1)}^2=n^2+2n+1$$$$\Rightarrow n+1=\sqrt{1+n(n+2)}$$$$\mathrm{If}\mathrm{we}\mathrm{put}2023\mathrm{as}\mathrm{the}\mathrm{value}\mathrm{of}n\mathrm{then}\mathrm{it}\mathrm{will}\mathrm{be}$$$$2024=\sqrt{1+\left(2023\right)\left(2025\right)}$$$$\mathrm{If}\mathrm{we}\mathrm{put}2024\mathrm{as}\mathrm{the}\mathrm{value}\mathrm{of}n\mathrm{then}\mathrm{it}\mathrm{will}\mathrm{be}$$$$2025=\sqrt{1+\left(2024\right)\left(2026\right)}$$$$\mathrm{So}\mathrm{we}\mathrm{can}\mathrm{say}$$$$2026=\sqrt{1+\left(2025\right)\left(2027\right)}$$$$\mathrm{Now}2025=\sqrt{1+2024\sqrt{1+\left(2025\right)\left(2027\right)}}$$$$\mathrm{Now}2024=\sqrt{1+2023\sqrt{1+2024\sqrt{1+2025\sqrt{...}...}}}$$$$\mathrm{Thus}\mathrm{this}\mathrm{will}\mathrm{continue}\mathrm{to}\mathrm{infinity}$$$$2027=\sqrt{1+\left(2026\right)\left(2028\right)}$$$$\mathbf{So}2024\mathbf{is}\mathbf{the}\mathbf{answer}.$$

Commented by TonyCWX08 last updated on 13/Apr/24

$$Radical$$

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