Long Sequences...

Use $\dbinom{n}{r} + \dbinom{n}{r-1} = \dbinom{n+1}{r}$ repeatedly. $D^r = \dbinom{n+1}{r+1} + 2 \dbinom{n+1}{r+2} + \dbinom{n+1}{r+3}$ $=\dbinom{n+2}{r+2} + \dbinom{n+1}{r+3}$ $= \dbinom{n+3}{r+3}$

$N^r = \dbinom{n+1}{r+1} + 3(\dbinom{n+1}{r+2} +\dbinom{n+1}{r+3}) + \dbinom{n+1}{r+4}$

$Middle Term = (1+2) \dbinom{n+2}{r+3}$ $= \dbinom{n+2}{r+2} + 2 (\dbinom{n+2}{r+3}) +\dbinom{n+2}{r+4}$ $= \dbinom{n+3}{r+3} + \dbinom{n+3}{r+4}$ $= \dbinom{n+4}{r+4}$

$\Rightarrow E = \frac{N^r}{D^r} = \frac{(n+4)!}{(n-r)! (r+4)!} \times \frac{(n-r)! (r+3)!}{(n+3)!}$ $= \frac{n + 4}{r + 4} = \frac{n + k}{r + k}$ $\Rightarrow \boxed{k = 4}$

First of all, it's $_{n}C_{r}$ , not $^{n}C_{r}$ . It's very confusing when you write $4^n C_{r+1}$ because it looks like you're saying $\binom{4^n}{r+1}$ . Second, you can also write a binomial coefficient in this way: $\binom{n}{r}$ by writing "\binom{n}{r}" in LaTeX.

Now as for the solution, we can use the Vandermonde Identity, which says that for any $n,p,q,r \ge 0$ we have: $\sum_{k=0}^{q} \binom{p}{q - k} \binom{n}{r+k} = \binom{n+p}{r+q}$ In the numerator, we have: $\binom{n}{r} + 4 \binom{n}{r+1} + 6 \binom{n}{r+2} + 4 \binom{n}{r+3} + \binom{n}{r+4}$ $= \binom{4}{4} \binom{n}{r} + \binom{4}{3} \binom{n}{r+1} + \binom{4}{2} \binom{n}{r+2} + \binom{4}{1} \binom{n}{r+3} + \binom{4}{0} \binom{n}{r+4}$ $= \binom{n+4}{r+4}$ (using the Vandermonde Identity with $p = 4, q = 4$ ).

Similarly, in the denominator we have:

$\binom{n}{r} + 3 \binom{n}{r+1} + 3 \binom{n}{r+2} + \binom{n}{r+3}$ $= \binom{3}{3} \binom{n}{r} + \binom{3}{2} \binom{n}{r+1} + \binom{3}{1} \binom{n}{r+2} + \binom{3}{0} \binom{n}{r+3}$ $= \binom{n+3}{r+3}$ (using the Vandermonde Identity with $p = 3, q = 3$ ).

So, the expression simplifies to:

$\frac{\binom{n+4}{r+4}}{\binom{n+3}{r+3}} = \frac{(n+4)!}{(n-r)!(r+4)!} × \frac{(n-r)!(r+3)!}{(n+3)!} = \frac{n+4}{r+4}$

Therefore $k = \boxed{4}$ .

put n=4 , r=0 in the problem n get the ans....or you can take any value of n and r but there difference must be greater than or equal to 4