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

[lim_(n→∞) (2.2^3 .2^5 .....2^(n−1) .3^2 .3^4 .....3^n )^(1/(n^2 +1)) ]^4 =?

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

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

s=1+3+5+...+n−1 n−1=1+(t−1)2 t=((n−2)/2)+1 t=(n/2) s=(n/4){2×1+((n/2)−1)2} =(n/4){2+n−2} =(n^2 /4) s_1 =2+4+...+n n=2+(t_1 −1)2 t_1 =(n/2) s_1 =(n/4){2×2+((n/2)−1)×2} s_1 =(n/4){4+n−2} s_1 =((n^2 +2n)/4) contd lim_(n→∞) {2^(n^2 /4) ×3^((n^2 +2n)/4) }^(4/(n^2 +1)) lim_(n→∞) {2^(n^2 /(n^2 +1)) ×3^((n^2 +2n)/(n^2 +1)) } lim_(n→∞) {2^(1/(1+((1 )/n^2 ))) ×3^((1+(2/n))/(1+(1/n^2 ))) } =2×3=6

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

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

Thank you sir.

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

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

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

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

everything is welcome from your side sir!:)

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

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