QUESTION 3 [5] \& 3.1. Write the first four terms of the expansion, \( (2 x+3)^{8} \)

We use the Binomial Theorem which states that \[ (a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k. \] Here, \(a=2x\), \(b=3\), and \(n=8\). Therefore, the general term is \[ T(k)=\binom{8}{k}(2x)^{8-k}3^k. \] We now calculate the first four terms (corresponding to \(k=0,1,2,3\)): 1. For \(k=0\): \[ T(0)=\binom{8}{0}(2x)^8\cdot 3^0=1\cdot (2x)^8\cdot 1=(2^8)x^8=256x^8. \] 2. For \(k=1\): \[ T(1)=\binom{8}{1}(2x)^7\cdot 3^1=8\cdot (2^7x^7)\cdot 3=8\cdot 128x^7\cdot 3. \] \[ 8\cdot 128\cdot 3=3072, \quad \text{so} \quad T(1)=3072x^7. \] 3. For \(k=2\): \[ T(2)=\binom{8}{2}(2x)^6\cdot 3^2. \] \[ \binom{8}{2}=28,\quad (2x)^6=2^6x^6=64x^6,\quad 3^2=9. \] \[ T(2)=28\cdot 64x^6\cdot 9=28\cdot 576x^6. \] \[ 28\cdot 576=16128, \quad \text{so} \quad T(2)=16128x^6. \] 4. For \(k=3\): \[ T(3)=\binom{8}{3}(2x)^5\cdot 3^3. \] \[ \binom{8}{3}=56,\quad (2x)^5=2^5x^5=32x^5,\quad 3^3=27. \] \[ T(3)=56\cdot 32x^5\cdot 27=56\cdot (32\cdot 27)x^5. \] \[ 32\cdot 27=864,\quad 56\cdot 864=48384, \quad \text{so} \quad T(3)=48384x^5. \] Thus, the first four terms of the expansion are \[ 256x^8,\quad 3072x^7,\quad 16128x^6,\quad 48384x^5. \]