To expand the binomial [tex]\( (11a + b)^6 \)[/tex] using the binomial theorem, we'll follow a systematic approach. The binomial theorem states that:
[tex]\[
(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k
\][/tex]
where [tex]\( \binom{n}{k} \)[/tex] (read as "n choose k") is the binomial coefficient, calculated as:
[tex]\[
\binom{n}{k} = \frac{n!}{k!(n-k)!}
\][/tex]
In our case, [tex]\( x = 11a \)[/tex], [tex]\( y = b \)[/tex], and [tex]\( n = 6 \)[/tex].
The general term in the expansion is:
[tex]\[
\binom{6}{k} (11a)^{6-k} b^k
\][/tex]
We'll expand this term-by-term from [tex]\( k = 0 \)[/tex] to [tex]\( k = 6 \)[/tex]:
Step-by-step Expansion:
1. For [tex]\( k = 0 \)[/tex]:
[tex]\[
\binom{6}{0} (11a)^{6-0} b^0 = 1 \cdot (11a)^6 \cdot 1 = 1771561a^6
\][/tex]
2. For [tex]\( k = 1 \)[/tex]:
[tex]\[
\binom{6}{1} (11a)^{6-1} b^1 = 6 \cdot (11a)^5 \cdot b = 6 \cdot 161051a^5 \cdot b = 966306a^5b
\][/tex]
3. For [tex]\( k = 2 \)[/tex]:
[tex]\[
\binom{6}{2} (11a)^{6-2} b^2 = 15 \cdot (11a)^4 \cdot b^2 = 15 \cdot 14641a^4 \cdot b^2 = 219615a^4b^2
\][/tex]
4. For [tex]\( k = 3 \)[/tex]:
[tex]\[
\binom{6}{3} (11a)^{6-3} b^3 = 20 \cdot (11a)^3 \cdot b^3 = 20 \cdot 1331a^3 \cdot b^3 = 26620a^3b^3
\][/tex]
5. For [tex]\( k = 4 \)[/tex]:
[tex]\[
\binom{6}{4} (11a)^{6-4} b^4 = 15 \cdot (11a)^2 \cdot b^4 = 15 \cdot 121a^2 \cdot b^4 = 1815a^2b^4
\][/tex]
6. For [tex]\( k = 5 \)[/tex]:
[tex]\[
\binom{6}{5} (11a)^{6-5} b^5 = 6 \cdot 11a \cdot b^5 = 6 \cdot 11a \cdot b^5 = 66ab^5
\][/tex]
7. For [tex]\( k = 6 \)[/tex]:
[tex]\[
\binom{6}{6} (11a)^{6-6} b^6 = 1 \cdot a^0 \cdot b^6 = b^6
\][/tex]
Combining all these terms, we get the expanded form:
[tex]\[
(11a + b)^6 = 1771561a^6 + 966306a^5b + 219615a^4b^2 + 26620a^3b^3 + 1815a^2b^4 + 66ab^5 + b^6
\][/tex]
This is the expansion of the binomial [tex]\( (11a + b)^6 \)[/tex] using the binomial theorem.
