In a certain test there are $n$ questions In this test $2^k$ students gave wrong answer to atleast $\left(n-k\right)$questions, Where $k=0,1,2,\mathrm{3....}n$ .If the total ways of wrong answer is 4095, then the value of $\frac{n}{4}=----$

The number of students answering at least r questions in correctly is $2^{n-r}$

\ the number of students answering exactly $r(1\le r\le n-1)$ questions in correctly is

$2^{n-r}-2^{n(r+1)}$

Also the number of students answering all question wrongly is 2 1 ° = so total number of wrong answer is

$1.\left(2^{n-1}{2}^{n-2}\right)+2(2^{n-2}-2^{n-3})+3(2^{n-3}-2^{n-4})+---+(n-1)(2^1-2^0)+n{.2}^0$

$2^{n-1}+2^{n-2}+2^{n-3}+----2+2^{°}=4095$

$2^n-1=4095\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{2}^n=40962^n=2^{12}n=12So\frac{n}{4}=\frac{12}{4}=3$