Question

If ²ⁿC₃ : ⁿC₃ = 10, then \(\frac{n^{2}+3n}{n^{2}-3n+4}\) is equal to

Answer 1

Correct Answer: 2

\(\frac{^{2n}C_{3}}{^{n}C_{3}}=10\Rightarrow \frac{2n.(2n-1).(2n-2)}{n.(n-1)(n-2)}\)
\(\Rightarrow \frac{(2n-1).2}{n-2}\)
\(\Rightarrow n=8\)
Therefore,  \(\frac{n^{2}+3n}{n^{2}-3n+4}\) = \(\frac{88}{44}\) = 2

The answer is 2

Answer 2

Given:
\[ \frac{{^{2n}C_3}}{{^nC_3}} = 10 \] Breaking down the formula: 
\[ \frac{\frac{(2n)!}{3!(2n - 3)!}}{\frac{n!}{3!(n - 3)!}} = \frac{(2n)! / (2n - 3)!}{n! / (n - 3)!} \] Simplifying:
\[ \frac{2n(2n - 1)(2n - 2)}{n(n - 1)(n - 2)} = 10 \] Further Simplification:
\[ \frac{4(2n - 1)}{n - 2} = 10 \] Solving:
\[ 4(2n - 1) = 10(n - 2) \\ 8n - 4 = 10n - 20 \\ 2n = 16 \\ n = 8 \] Finding the ratio:
\[ \text{Ratio} = \frac{n^2 + 3n}{n^2 - 3n + 4} \] Substituting \( n = 8 \): \[ = \frac{8^2 + 3(8)}{8^2 - 3(8) + 4} = \frac{64 + 24}{64 - 24 + 4} = \frac{88}{44} = 2 \]