Question

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

• A. 1
• B. 2
• C. 4
• D. 5

Answer 1

The correct option is C 4

Use ${}^n{C}_r{+}^n{C}_{r-1}{=}^{n+1}{C}_r$ repeatedly.

Numerator

${}^n{C}_r+4^n{C}_{r+1}+6^n{C}_{r+2}+4^n{C}_{r+3}+n{C}_{r+4}$

$={(}^n{C}_r{+}^n{C}_{r+1})+3{(}^n{C}_{r+1}{+}^n{C}_{r+2})+3{(}^n{C}_{r+2}{+}^n{C}_{r+3})+{(}^n{C}_{r+3}{+}^n{C}_{r+4})$

$={(}^{n+1}{C}_{r+1})+3{(}^{n+1}{C}_{r+2})+3{(}^{n+1}{C}_{r+3})+{(}^{n+1}{C}_{r+4})$

$={(}^{n+1}{C}_{r+1}{+}^{n+1}{C}_{r+2})+2{(}^{n+1}{C}_{r+2}{+}^{n+1}{C}_{r+3})+{(}^{n+1}{C}_{r+3}{+}^{n+1}{C}_{r+4})$

$={(}^{n+2}{C}_{r+2})+2{(}^{n+2}{C}_{r+3})+{(}^{n+2}{C}_{r+4})$

$={(}^{n+2}{C}_{r+2}{+}^{n+2}{C}_{r+3})+{(}^{n+2}{C}_{r+3}{+}^{n+2}{C}_{r+4})$

${=}^{n+3}{C}_{r+3}{+}^{n+3}{C}_{r+4}$

${=}^{n+4}{C}_{r+4}$

Similarly, Denominator

${}^n{C}_r+3^n{C}_{r+1}+3^n{C}_{r+2}{+}^n{C}_{r+3}$

${=}^{n+3}{C}_{r+3}$

$\therefore \frac{{}^n{C}_r+4^n{C}_{r+1}+6^n{C}_{r+2}+4^n{C}_{r+3}{+}^n{C}_{r+4}}{{}^n{C}_r+3^n{C}_{r+1}+3^n{C}_{r+2}{+}^n{C}_{r+3}}=\frac{{}^{n+4}{C}_{r+4}}{{}^{n+3}{C}_{r+3}}$

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

$\therefore k=4$