Question

d) $x^6 + 2x^5 + 4x^4 + 8x^3 + 16x^2 + 32x + 64$

1. expand $(a + 4)^5$.

posted mon dec 8, 2025 at 11:24 am

Explanation:

Step1: Recall the Binomial Theorem

The binomial theorem states that \((x + y)^n=\sum_{k = 0}^{n}\binom{n}{k}x^{n - k}y^{k}\), where \(\binom{n}{k}=\frac{n!}{k!(n - k)!}\) and \(n!=n\times(n - 1)\times\cdots\times1\) for \(n\geq1\) and \(0!=1\). For \((a + 4)^5\), we have \(x = a\), \(y = 4\) and \(n = 5\).

Step2: Calculate each term for \(k = 0\) to \(k = 5\)

• When \(k = 0\):

\(\binom{5}{0}a^{5-0}\times4^{0}=\frac{5!}{0!(5 - 0)!}a^{5}\times1 = 1\times a^{5}\times1=a^{5}\)

• When \(k = 1\):

\(\binom{5}{1}a^{5 - 1}\times4^{1}=\frac{5!}{1!(5 - 1)!}a^{4}\times4=\frac{5\times4!}{1\times4!}a^{4}\times4 = 5\times4a^{4}=20a^{4}\)

• When \(k = 2\):

\(\binom{5}{2}a^{5 - 2}\times4^{2}=\frac{5!}{2!(5 - 2)!}a^{3}\times16=\frac{5\times4\times3!}{2\times1\times3!}a^{3}\times16 = 10\times16a^{3}=160a^{3}\)

• When \(k = 3\):

\(\binom{5}{3}a^{5 - 3}\times4^{3}=\frac{5!}{3!(5 - 3)!}a^{2}\times64=\frac{5\times4\times3!}{3!\times2\times1}a^{2}\times64 = 10\times64a^{2}=640a^{2}\)

• When \(k = 4\):

\(\binom{5}{4}a^{5 - 4}\times4^{4}=\frac{5!}{4!(5 - 4)!}a^{1}\times256=\frac{5\times4!}{4!\times1}a\times256 = 5\times256a = 1280a\)

• When \(k = 5\):

\(\binom{5}{5}a^{5 - 5}\times4^{5}=\frac{5!}{5!(5 - 5)!}a^{0}\times1024=1\times1\times1024 = 1024\)

Step3: Sum up all the terms

\((a + 4)^5=a^{5}+20a^{4}+160a^{3}+640a^{2}+1280a + 1024\)

Answer:

\(a^{5}+20a^{4}+160a^{3}+640a^{2}+1280a + 1024\)