Question Number 189501 by mathlove last updated on 18/Mar/23
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{x}+\mathrm{2}{x}+\mathrm{3}{x}+\mathrm{4}{x}+.....+{nx}} −\mathrm{1}}{{x}}=? \\ $$
Answered by mehdee42 last updated on 18/Mar/23
$${hop}\rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \left(\mathrm{1}+\mathrm{2}+...+{n}\right){e}^{{x}+\mathrm{2}{x}+...+{nx}} =\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$
Commented by mathlove last updated on 18/Mar/23
$${with}\:{out}\:{Hopetal}\:{Rul} \\ $$
Answered by mr W last updated on 18/Mar/23
![e^(x+2x+3x+...+nx) =e^((n(n+1)x)/2) =1+((n(n+1)x)/2)+((n^2 (n+1)^2 x^2 )/(2^2 2!))+... lim_(x→0) ((e^(x+2x+3x+4x+.....+nx) −1)/x) =lim_(x→0) [((n(n+1))/2)+((n^2 (n+1)^2 x)/(2^2 2!))+...] =((n(n+1))/2)](Q189541.png)
$${e}^{{x}+\mathrm{2}{x}+\mathrm{3}{x}+...+{nx}} ={e}^{\frac{{n}\left({n}+\mathrm{1}\right){x}}{\mathrm{2}}} =\mathrm{1}+\frac{{n}\left({n}+\mathrm{1}\right){x}}{\mathrm{2}}+\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} \mathrm{2}!}+... \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{x}+\mathrm{2}{x}+\mathrm{3}{x}+\mathrm{4}{x}+.....+{nx}} −\mathrm{1}}{{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}+\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} {x}}{\mathrm{2}^{\mathrm{2}} \mathrm{2}!}+...\right] \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$
Commented by mathlove last updated on 18/Mar/23
$${thanks}\:{dear} \\ $$
Answered by CElcedricjunior last updated on 18/Mar/23
$$\boldsymbol{{l}}=\frac{\boldsymbol{{n}}\left(\boldsymbol{{n}}+\mathrm{1}\right)}{\mathrm{2}} \\ $$