lim-n-2-2-3-2-5-2-n-1-3-2-3-4-3-n-1-n-2-1-4-

Question Number 36969 by rahul 19 last updated on 07/Jun/18

$$\left[\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{2}.\mathrm{2}^{\mathrm{3}} .\mathrm{2}^{\mathrm{5}} …..\mathrm{2}^{\mathrm{n}−\mathrm{1}} .\mathrm{3}^{\mathrm{2}} .\mathrm{3}^{\mathrm{4}} …..\mathrm{3}^{\mathrm{n}} \right)^{\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} +\mathrm{1}}} \right]^{\mathrm{4}} =? \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 07/Jun/18

$${s}=\mathrm{1}+\mathrm{3}+\mathrm{5}+…+{n}−\mathrm{1} \\ $$$${n}−\mathrm{1}=\mathrm{1}+\left({t}−\mathrm{1}\right)\mathrm{2} \\ $$$${t}=\frac{{n}−\mathrm{2}}{\mathrm{2}}+\mathrm{1} \\ $$$${t}=\frac{{n}}{\mathrm{2}} \\ $$$${s}=\frac{{n}}{\mathrm{4}}\left\{\mathrm{2}×\mathrm{1}+\left(\frac{{n}}{\mathrm{2}}−\mathrm{1}\right)\mathrm{2}\right\} \\ $$$$=\frac{{n}}{\mathrm{4}}\left\{\mathrm{2}+{n}−\mathrm{2}\right\} \\ $$$$=\frac{{n}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${s}_{\mathrm{1}} =\mathrm{2}+\mathrm{4}+…+{n} \\ $$$${n}=\mathrm{2}+\left({t}_{\mathrm{1}} −\mathrm{1}\right)\mathrm{2} \\ $$$${t}_{\mathrm{1}} =\frac{{n}}{\mathrm{2}} \\ $$$${s}_{\mathrm{1}} =\frac{{n}}{\mathrm{4}}\left\{\mathrm{2}×\mathrm{2}+\left(\frac{{n}}{\mathrm{2}}−\mathrm{1}\right)×\mathrm{2}\right\} \\ $$$${s}_{\mathrm{1}} =\frac{{n}}{\mathrm{4}}\left\{\mathrm{4}+{n}−\mathrm{2}\right\} \\ $$$${s}_{\mathrm{1}} =\frac{{n}^{\mathrm{2}} +\mathrm{2}{n}}{\mathrm{4}} \\ $$$${contd} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{\mathrm{2}^{\frac{{n}^{\mathrm{2}} }{\mathrm{4}}} ×\mathrm{3}^{\frac{{n}^{\mathrm{2}} +\mathrm{2}{n}}{\mathrm{4}}} \:\:\right\}^{\frac{\mathrm{4}}{{n}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{\mathrm{2}^{\frac{{n}^{\mathrm{2}} }{{n}^{\mathrm{2}} +\mathrm{1}}} ×\mathrm{3}^{\frac{{n}^{\mathrm{2}} +\mathrm{2}{n}}{{n}^{\mathrm{2}} +\mathrm{1}}} \right\} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}\:}{{n}^{\mathrm{2}} }}} ×\mathrm{3}^{\frac{\mathrm{1}+\frac{\mathrm{2}}{{n}}}{\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }}} \:\right\} \\ $$$$=\mathrm{2}×\mathrm{3}=\mathrm{6} \\ $$$$ \\ $$

Commented by rahul 19 last updated on 07/Jun/18

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 07/Jun/18

$${i}\:{want}\:{to}\:{share}\:{that}\:{newton}\:{laws}\:{originated} \\ $$$${in}\:{india}\:{do}\:{i}\:{share}\:{kn}\:{public}\:{domain}\:{pls}\:{reply} \\ $$

Commented by rahul 19 last updated on 07/Jun/18

$$\left.\mathrm{everything}\:\mathrm{is}\:\mathrm{welcome}\:\mathrm{from}\:\mathrm{your}\:\mathrm{side}\:\mathrm{sir}!:\right) \\ $$

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