Question

If $\frac{{{}^{n}C}_r+4{{}^{n}C}_{r+1}+6{{}^{n}C}_{r+2}+4{{}^{n}C}_{r+3}+{{}^{n}C}_{r+4}}{{{}^{n}C}_r+3{{}^{n}C}_{r+1}+3{{}^{n}C}_{r+2}+{{}^{n}C}_{r+3}}=\frac{n+k}{r+k}$. Find the value of k

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

The correct option is B 4

$\frac{{{}^{n}C}_r+4{{}^{n}C}_{r+1}+6{{}^{n}C}_{r+2}+4{{}^{n}C}_{r+3}+{{}^{n}C}_{r+4}}{{{}^{n}C}_r+3{{}^{n}C}_{r+1}+3{{}^{n}C}_{r+2}+{{}^{n}C}_{r+3}}$
$=1+\frac{{{}^{n}C}_{r+1}+3{{}^{n}C}_{r+2}+3{{}^{n}C}_{r+3}+{{}^{n}C}_{r+4}}{{{}^{n}C}_r+3{{}^{n}C}_{r+1}+3{{}^{n}C}_{r+2}+{{}^{n}C}_{r+3}}$
using ${{}^{n}C}_r+{{}^{n}C}_{r-1}={{}^{n+1}C}_r$
$=1+\frac{{{}^{n+1}C}_{r+2}+2{{}^{n+1}C}_{r+3}+{{}^{n+1}C}_{r+4}}{{{}^{n+1}C}_{r+1}+2{{}^{n+1}C}_{r+2}+{{}^{n+1}C}_{r+3}}$
$=1+\frac{{{}^{n+2}C}_{r+3}+{{}^{n+2}C}_{r+4}}{{{}^{n+2}C}_{r+2}+{{}^{n+2}C}_{r+3}}$
$=1+\frac{{{}^{n+3}C}_{r+4}}{{{}^{n+3}C}_{r+3}}$
$=\frac{n+4}{r+4}$
$\therefore k=4$
Hence, option B.