Question

expand ((5x + 4y)^6) using pascal’s triangle.
((5x + 4y)^6 = square)
(simplify your answer.)

Explanation:

Step1: Recall Pascal's Triangle for \(n = 6\)

The coefficients for \((a + b)^6\) from Pascal's Triangle are \(1, 6, 15, 20, 15, 6, 1\).

Step2: Apply the Binomial Theorem

The binomial expansion of \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\). For \((5x+4y)^6\), \(a = 5x\), \(b = 4y\), \(n = 6\).

• For \(k = 0\): \(\binom{6}{0}(5x)^{6}(4y)^{0}=1\times15625x^{6}\times1 = 15625x^{6}\)
• For \(k = 1\): \(\binom{6}{1}(5x)^{5}(4y)^{1}=6\times3125x^{5}\times4y = 6\times12500x^{5}y=75000x^{5}y\)
• For \(k = 2\): \(\binom{6}{2}(5x)^{4}(4y)^{2}=15\times625x^{4}\times16y^{2}=15\times10000x^{4}y^{2}=150000x^{4}y^{2}\)
• For \(k = 3\): \(\binom{6}{3}(5x)^{3}(4y)^{3}=20\times125x^{3}\times64y^{3}=20\times8000x^{3}y^{3}=160000x^{3}y^{3}\)
• For \(k = 4\): \(\binom{6}{4}(5x)^{2}(4y)^{4}=15\times25x^{2}\times256y^{4}=15\times6400x^{2}y^{4}=96000x^{2}y^{4}\)
• For \(k = 5\): \(\binom{6}{5}(5x)^{1}(4y)^{5}=6\times5x\times1024y^{5}=6\times5120xy^{5}=30720xy^{5}\)
• For \(k = 6\): \(\binom{6}{6}(5x)^{0}(4y)^{6}=1\times1\times4096y^{6}=4096y^{6}\)

Step3: Sum all terms

\((5x + 4y)^6=15625x^{6}+75000x^{5}y + 150000x^{4}y^{2}+160000x^{3}y^{3}+96000x^{2}y^{4}+30720xy^{5}+4096y^{6}\)

Answer:

\(15625x^{6}+75000x^{5}y + 150000x^{4}y^{2}+160000x^{3}y^{3}+96000x^{2}y^{4}+30720xy^{5}+4096y^{6}\)