Question Number 145304 by imjagoll last updated on 04/Jul/21
$$\:\mathrm{1}+\frac{\mathrm{3x}}{\mathrm{1}!}\:+\frac{\mathrm{5x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{\mathrm{7x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{\mathrm{9x}^{\mathrm{4}} }{\mathrm{4}!}+…+\infty=? \\ $$
Answered by qaz last updated on 04/Jul/21
$$\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{2n}+\mathrm{1}\right)\mathrm{x}^{\mathrm{n}} }{\mathrm{n}!}=\left(\mathrm{2xD}+\mathrm{1}\right)\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{n}!}=\left(\mathrm{2xD}+\mathrm{1}\right)\mathrm{e}^{\mathrm{x}} =\left(\mathrm{2x}+\mathrm{1}\right)\mathrm{e}^{\mathrm{x}} \\ $$