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Question Number 6739 by Tawakalitu. last updated on 20/Jul/16

Find the 102rd term of f(x) = e^(2x^3 ) at x = 0

$${Find}\:{the}\:\mathrm{102}{rd}\:{term}\:{of}\:\:\:{f}\left({x}\right)\:=\:{e}^{\mathrm{2}{x}^{\mathrm{3}} } \:\:\:{at}\:\:{x}\:=\:\mathrm{0} \\ $$

Answered by Yozzii last updated on 20/Jul/16

f(u)=e^u =Σ_(n=0) ^∞ (u^n /(n!)) ⇒T(n+1)=(u^n /(n!)) (n∈N+{0}) n+1=102⇒n=101 ∴T(102)=(u^(101) /(101!)) Let u=2x^3 . ∴ T(102)=((2^(101) x^(303) )/(101!)) At x=0, T(102)=0. −−−−−−−−−−−−−−−−−−−−−−−− 101!=1×2×3×4×5×6×...×98×99×100×101 101!=(Π_(i=1) ^(51) (2i−1))(Π_(i=1) ^(50) 2i) 101!=2^(50) ×50!×Π_(i=1) ^(51) (2i−1) 101!=2^(50) ×2^(25) ×25!×Π_(i=1) ^(25) (2i−1)×Π_(i=1) ^(51) (2i−1) ... omit this scribbling.

$${f}\left({u}\right)={e}^{{u}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{u}^{{n}} }{{n}!} \\ $$$$\Rightarrow{T}\left({n}+\mathrm{1}\right)=\frac{{u}^{{n}} }{{n}!}\:\:\:\left({n}\in\mathbb{N}+\left\{\mathrm{0}\right\}\right) \\ $$$${n}+\mathrm{1}=\mathrm{102}\Rightarrow{n}=\mathrm{101} \\ $$$$\therefore{T}\left(\mathrm{102}\right)=\frac{{u}^{\mathrm{101}} }{\mathrm{101}!} \\ $$$${Let}\:{u}=\mathrm{2}{x}^{\mathrm{3}} .\:\therefore\:{T}\left(\mathrm{102}\right)=\frac{\mathrm{2}^{\mathrm{101}} {x}^{\mathrm{303}} }{\mathrm{101}!} \\ $$$${At}\:{x}=\mathrm{0},\:{T}\left(\mathrm{102}\right)=\mathrm{0}. \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$$\mathrm{101}!=\mathrm{1}×\mathrm{2}×\mathrm{3}×\mathrm{4}×\mathrm{5}×\mathrm{6}×...×\mathrm{98}×\mathrm{99}×\mathrm{100}×\mathrm{101} \\ $$$$\mathrm{101}!=\left(\underset{{i}=\mathrm{1}} {\overset{\mathrm{51}} {\prod}}\left(\mathrm{2}{i}−\mathrm{1}\right)\right)\left(\underset{{i}=\mathrm{1}} {\overset{\mathrm{50}} {\prod}}\mathrm{2}{i}\right) \\ $$$$\mathrm{101}!=\mathrm{2}^{\mathrm{50}} ×\mathrm{50}!×\underset{{i}=\mathrm{1}} {\overset{\mathrm{51}} {\prod}}\left(\mathrm{2}{i}−\mathrm{1}\right) \\ $$$$\mathrm{101}!=\mathrm{2}^{\mathrm{50}} ×\mathrm{2}^{\mathrm{25}} ×\mathrm{25}!×\underset{{i}=\mathrm{1}} {\overset{\mathrm{25}} {\prod}}\left(\mathrm{2}{i}−\mathrm{1}\right)×\underset{{i}=\mathrm{1}} {\overset{\mathrm{51}} {\prod}}\left(\mathrm{2}{i}−\mathrm{1}\right) \\ $$$$...\:{omit}\:{this}\:{scribbling}. \\ $$$$ \\ $$

Commented by Tawakalitu. last updated on 20/Jul/16

Thanks so much

$${Thanks}\:{so}\:{much} \\ $$

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