Question

1. use the binomial theorem expression to find the expansion of ((2x + 1)^4).

((2x + 1)^4 = x^4 + 4x^3 + 6x^2 + 4x + 1)
((2x + 1)^4 = 32x^4 + 8x^3 + 16x^2 + x + 8)
((2x + 1)^4 = x^4 + x^3 + x^2 + x)
((2x + 1)^4 = 16x^4 + 32x^3 + 24x^2 + 8x + 1)

Explanation:

Step1: Recall binomial theorem

The binomial theorem for $(a+b)^n$ is $\sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^{k}$, where $\binom{n}{k}=\frac{n!}{k!(n-k)!}$. For $(2x+1)^4$, $a=2x$, $b=1$, $n=4$.

Step2: Calculate term $k=0$

$\binom{4}{0}(2x)^{4-0}(1)^0 = 1 \cdot (2x)^4 \cdot 1 = 16x^4$

Step3: Calculate term $k=1$

$\binom{4}{1}(2x)^{4-1}(1)^1 = 4 \cdot (2x)^3 \cdot 1 = 4 \cdot 8x^3 = 32x^3$

Step4: Calculate term $k=2$

$\binom{4}{2}(2x)^{4-2}(1)^2 = 6 \cdot (2x)^2 \cdot 1 = 6 \cdot 4x^2 = 24x^2$

Step5: Calculate term $k=3$

$\binom{4}{3}(2x)^{4-3}(1)^3 = 4 \cdot (2x)^1 \cdot 1 = 4 \cdot 2x = 8x$

Step6: Calculate term $k=4$

$\binom{4}{4}(2x)^{4-4}(1)^4 = 1 \cdot (2x)^0 \cdot 1 = 1$

Step7: Sum all terms

Add the calculated terms together: $16x^4 + 32x^3 + 24x^2 + 8x + 1$

Answer:

$\boldsymbol{(2x + 1)^4 = 16x^4 + 32x^3 + 24x^2 + 8x + 1}$