1. **State the problem:** We are given the first three terms in ascending powers of $x$ in the expansion of $$(3+px)^n$$ as $$243 + 810x + qx^2$$. We need to find the values of $n$, $p$, and $q$. 2. **Recall the binomial expansion formula:** $$ (a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k $$ The first three terms correspond to $k=0,1,2$: $$ T_0 = \binom{n}{0} 3^n (px)^0 = 3^n $$ $$ T_1 = \binom{n}{1} 3^{n-1} (px)^1 = n \cdot 3^{n-1} p x $$ $$ T_2 = \binom{n}{2} 3^{n-2} (px)^2 = \frac{n(n-1)}{2} 3^{n-2} p^2 x^2 $$ 3. **Match the given terms:** - Constant term: $$3^n = 243$$ - Coefficient of $x$: $$n \cdot 3^{n-1} p = 810$$ - Coefficient of $x^2$: $$\frac{n(n-1)}{2} 3^{n-2} p^2 = q$$ 4. **Solve for $n$:** Since $$3^n = 243$$ and $$243 = 3^5$$, we get $$n = 5$$. 5. **Substitute $n=5$ into the $x$ coefficient equation:** $$5 \cdot 3^{4} p = 810$$ Calculate $$3^4 = 81$$: $$5 \cdot 81 \cdot p = 810$$ $$405p = 810$$ Divide both sides by 405: $$\cancel{405}p = \frac{810}{\cancel{405}}$$ $$p = 2$$ 6. **Find $q$ using $n=5$ and $p=2$:** $$q = \frac{5 \cdot 4}{2} 3^{3} (2)^2$$ Calculate each part: $$\frac{20}{2} = 10$$ $$3^3 = 27$$ $$(2)^2 = 4$$ Multiply: $$q = 10 \times 27 \times 4 = 1080$$ **Final answers:** $$n = 5, \quad p = 2, \quad q = 1080$$