Invincible e

Note first that

$e^{x} = \displaystyle\sum_{k=0}^{\infty} \dfrac{x^{k}}{k!} = 1 + x + \dfrac{x^{2}}{2} + \dfrac{x^{3}}{6} + .... .$

Since we are looking for the coefficient of $x^{2}$ in $(1 + e^{x})^{10}$ , we only need to involve the first three terms from the power series expansion of $e^{x}$ . Thus we are looking for the terms involving $x^{2}$ in $(2 + x + \frac{x^{2}}{2})^{10}$ , which will be

$\dbinom{10}{2}(x)^{2}*2^{8} + \dbinom{10}{1}(\dfrac{x^{2}}{2})*2^{9} = (45*2^{8} + 10*2^{8})x^{2} = 14080x^{2}$ .

Thus the desired coefficient is $\boxed{14080}$ .

u can do that for small n like 10..but if huge use this method

assume

(1+e^x)^10=A+Bx+Cx^2+Dx^3...

differntiate w.r.t x both sides 2 times to get

10e^x (e^x+1)^8 (10e^x+1) = 2C+6Dx..........terms in x

just sub x=0

110*2^8 = 2C

C=110*2^7=14080.......ANS

cheers

Expand using Binomial Theorem ,

$(1+e^{x})^{10} = 1+{10\choose 1} e^x+{10\choose 2} e^{2x}+{10\choose 3}e^{3x}+...................+{10\choose 10}e^{10x}$

Now,

$e^{mx}=1+mx+\frac{m^2x^2}{2}+\frac{m^3x^3}{3!}+................$

So coefficient of $x^2$ is,

$\sum \limits_{i=1}^{10}{10\choose i} \frac{i^2}{2}=5+90+540+1680+3150+3780+2940+1440+405+50=\fbox{14080}$

By the Maclaurin 's series, use the binomial theorem first. After which find the expansion for e^x and slowly manipulate the coefficients. I will be giving a more detailed solution soon, but that was just a summary.