We have, $x^2+1-2x=0$
$\Rightarrow (x-1)^2=0$
$\Rightarrow x=1$
$\therefore x^{99}+\frac {1}{x^{99}}=1+1=2$
If the sum of the coefficients in the expansion of $(x+y)^n$ is 4096, then the greatest coefficient in the expansion is _ _ _ _ . Solution $C_0 + C_1 + C_2 + C_3 + ......................... + C_n =4096 $ $\therefore 2^n = 4096 =2^{12} $ $\Rightarrow n = 12 $ Greatest coefficient = ${}^{12}{C_6} = 924$