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Let $\mathrm{z}=(1+\mathrm{i})(1+2 \mathrm{i})(1+3 \mathrm{i}) \ldots(1+\mathrm{ni})$, where $\mathrm{i}=\sqrt{-1}$. If $|\mathrm{z}|^{2}=44200$, then n is equal to

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  • 22 October, 2026 at 6:00 pm #12294
    UPLIFTJEE
    Keymaster

    Let $\mathrm{z}=(1+\mathrm{i})(1+2 \mathrm{i})(1+3 \mathrm{i}) \ldots(1+\mathrm{ni})$, where $\mathrm{i}=\sqrt{-1}$. If $|\mathrm{z}|^{2}=44200$, then n is equal to [JEE MAIN 24 Jan 2026 S II]

    22 October, 2026 at 6:00 pm #12295
    UPLIFTJEE
    Keymaster

    Solution

    $|\mathrm{z}|=|1+\mathrm{i}\|1+2 \mathrm{i}\| 1+3 \mathrm{i}| \cdots \cdots|1+\mathrm{ni}| \Rightarrow|\mathrm{z}|^{2}=(1+1)(1+4)(1+9) \ldots\left(1+\mathrm{n}^{2}\right)$

    \[ |z|^{2}=44200=2 \times 5 \times 10 \times 17 \times 26 \Rightarrow|z|^{2}=(1+1)\left(1+2^{2}\right)\left(1+3^{2}\right)\left(1+4^{2}\right)\left(1+5^{2}\right) \]

    From (1) $\therefore \mathrm{n}=5$

    22 October, 2026 at 6:00 pm #12296
    UPLIFTJEE
    Keymaster

    Answer

    5

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