Find Index Value for Consecutive Binomial Expansion Coefficients Ratio

To find the value of $n$, let the coefficients of three consecutive terms in the expansion of $(1 + x)^{N}$, where $N = n + 5$, be $\binom{N}{r-1}$, $\binom{N}{r}$, and $\binom{N}{r+1}$.

According to the given condition, these coefficients are in the ratio $5 : 10 : 14$: $\binom{N}{r-1} : \binom{N}{r} : \binom{N}{r+1} = 5 : 10 : 14$

Taking the ratio of the second term to the first term: $\frac{\binom{N}{r}}{\binom{N}{r-1}} = \frac{10}{5} = 2$

Using the identity $\frac{\binom{N}{r}}{\binom{N}{r-1}} = \frac{N - r + 1}{r}$, we get: $\frac{N - r + 1}{r} = 2$ $N - r + 1 = 2r$ $N + 1 = 3r \quad \text{--- (Equation 1)}$

Next, taking the ratio of the third term to the second term: $\frac{\binom{N}{r+1}}{\binom{N}{r}} = \frac{14}{10} = \frac{7}{5}$

Using the identity $\frac{\binom{N}{r+1}}{\binom{N}{r}} = \frac{N - r}{r + 1}$, we get: $\frac{N - r}{r + 1} = \frac{7}{5}$ $5(N - r) = 7(r + 1)$ $5N - 5r = 7r + 7$ $5N - 12r = 7 \quad \text{--- (Equation 2)}$

From Equation 1, we express $r$ in terms of $N$: $r = \frac{N + 1}{3}$

Substituting this value of $r$ into Equation 2: $5N - 12\left(\frac{N + 1}{3}\right) = 7$ $5N - 4(N + 1) = 7$ $5N - 4N - 4 = 7$ $N = 11$

Since $N = n + 5$: $n + 5 = 11 \implies n = 6$