Question

use the binomial theorem to expand ((2a + 2b)^6).
a. (64a^5 + 80a^4b + 320a^3b^2 + 320a^2b^3 + 80ab^4 + 10b^5)
b. (64a^6 + 384a^5b + 960a^4b^2 + 1280a^3b^3 + 960a^2b^4 + 384ab^5 + 64b^6)
c. (64a^6 + 160a^4b + 640a^3b^2 + 160ab^4 + 32b^5)
d. (64a^6 + 160a^4b + 320a^3b^2 + 320a^2b^3 + 160ab^4 + 20b^5)

Explanation:

Step1: Recall Binomial Theorem

For $(x+y)^n=\sum_{k=0}^{n}\binom{n}{k}x^{n-k}y^{k}$

Step2: Identify $x,y,n$

Here $x=2a$, $y=2b$, $n=6$

Step3: Calculate each term

• $k=0$: $\binom{6}{0}(2a)^6(2b)^0=1\cdot64a^6\cdot1=64a^6$
• $k=1$: $\binom{6}{1}(2a)^5(2b)^1=6\cdot32a^5\cdot2b=384a^5b$
• $k=2$: $\binom{6}{2}(2a)^4(2b)^2=15\cdot16a^4\cdot4b^2=960a^4b^2$
• $k=3$: $\binom{6}{3}(2a)^3(2b)^3=20\cdot8a^3\cdot8b^3=1280a^3b^3$
• $k=4$: $\binom{6}{4}(2a)^2(2b)^4=15\cdot4a^2\cdot16b^4=960a^2b^4$
• $k=5$: $\binom{6}{5}(2a)^1(2b)^5=6\cdot2a\cdot32b^5=384ab^5$
• $k=6$: $\binom{6}{6}(2a)^0(2b)^6=1\cdot1\cdot64b^6=64b^6$

Step4: Combine all terms

Sum the terms to get the expanded form.

Answer:

B. $64a^{6} + 384a^{5}b + 960a^{4}b^{2} + 1280a^{3}b^{3} + 960a^{2}b^{4} + 384ab^{5} + 64b^{6}$