Question:

Let the general term in the expansion of $(a+bx)^n$ be $T_{k+1} = \binom{n}{k} a^{n-k} (bx)^k$. The coefficient of $x^k$ is $C_k = \binom{n}{k} a^{n-k} b^k$. For the expansion of $(3+7x)^{29}$, we have $a=3$, $b=7$, and $n=29$. The coefficient of $x^r$ is $C_r = \binom{29}{r} 3^{29-r} 7^r$. The coefficient of $x^{r+1}$ is $C_{r+1} = \binom{29}{r+1} 3^{29-(r+1)} 7^{r+1} = \binom{29}{r+1} 3^{28-r} 7^{r+1}$. Given that the coefficients of $x^r$ and $x^{r+1}$ are equal, we set $C_r = C_{r+1}$: $$ \binom{29}{r} 3^{29-r} 7^r = \binom{29}{r+1} 3^{28-r} 7^{r+1} $$ Using the definition of binomial coefficients $\binom{n}{k} = \frac{n!}{k!(n-k)!}$: $$ \frac{29!}{r!(29-r)!} 3^{29-r} 7^r = \frac{29!}{(r+1)!(29-(r+1))!} 3^{28-r} 7^{r+1} $$ $$ \frac{29!}{r!(29-r)!} 3^{29-r} 7^r = \frac{29!}{(r+1)!(28-r)!} 3^{28-r} 7^{r+1} $$ To simplify, we can divide both sides by common factors: $$ \frac{3^{29-r} 7^r}{r!(29-r)!} = \frac{3^{28-r} 7^{r+1}}{(r+1)!(28-r)!} $$ Rearrange the terms to group similar bases and factorials: $$ \frac{3^{29-r}}{3^{28-r}} \cdot \frac{7^r}{7^{r+1}} = \frac{r!(29-r)!}{(r+1)!(28-r)!} $$ Simplify the exponential terms and expand the factorials: $$ 3^{29-r-(28-r)} \cdot 7^{r-(r+1)} = \frac{r! \cdot (29-r) \cdot (28-r)!}{(r+1) \cdot r! \cdot (28-r)!} $$ $$ 3^1 \cdot 7^{-1} = \frac{29-r}{r+1} $$ $$ \frac{3}{7} = \frac{29-r}{r+1} $$ Now, solve for $r$: $$ 3(r+1) = 7(29-r) $$ $$ 3r + 3 = 203 - 7r $$ $$ 3r + 7r = 203 - 3 $$ $$ 10r = 200 $$ $$ r = 20 $$