1. **Problem statement:** In the binomial expansion of $ (1 + kx)^6 $, the coefficient of $ x^3 $ is twice the coefficient of $ x^2 $. Find $ k $. 2. **Formula:** The general term in the expansion of $ (1 + kx)^6 $ is given by $$ T_{r+1} = \binom{6}{r} (1)^{6-r} (kx)^r = \binom{6}{r} k^r x^r $$ where $ r = 0, 1, 2, \ldots, 6 $. 3. **Coefficients:** - Coefficient of $ x^2 $ is $ \binom{6}{2} k^2 = 15 k^2 $ - Coefficient of $ x^3 $ is $ \binom{6}{3} k^3 = 20 k^3 $ 4. **Condition given:** Coefficient of $ x^3 $ is twice coefficient of $ x^2 $: $$ 20 k^3 = 2 \times 15 k^2 $$ 5. **Solve for $ k $:** $$ 20 k^3 = 30 k^2 $$ Divide both sides by $ k^2 $ (assuming $ k \neq 0 $): $$ 20 k = 30 $$ $$ k = \frac{30}{20} = \frac{3}{2} = 1.5 $$ --- 1. **Problem statement:** The coefficient of $ x^4 $ in the expansion of $ (a + x)^6 $ is 1600. Find the two possible values of $ a $. 2. **Formula:** The general term in the expansion of $ (a + x)^6 $ is $$ T_{r+1} = \binom{6}{r} a^{6-r} x^r $$ 3. **Coefficient of $ x^4 $:** $$ \binom{6}{4} a^{6-4} = \binom{6}{4} a^2 = 15 a^2 $$ 4. **Given:** $$ 15 a^2 = 1600 $$ 5. **Solve for $ a $:** $$ a^2 = \frac{1600}{15} = \frac{320}{3} $$ $$ a = \pm \sqrt{\frac{320}{3}} = \pm \frac{8 \sqrt{15}}{3} $$ **Final answers:** - $ k = 1.5 $ - $ a = \pm \frac{8 \sqrt{15}}{3} $