To find the coefficient of the seventh term in the expansion of [tex]\((x + 2y)^{10}\)[/tex], we can use the binomial theorem.
The binomial theorem states that for any positive integer [tex]\(n\)[/tex],
[tex]\[
(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k
\][/tex]
where [tex]\(\binom{n}{k}\)[/tex] is the binomial coefficient given by
[tex]\[
\binom{n}{k} = \frac{n!}{k!(n-k)!}
\][/tex]
To find the seventh term in the expansion of [tex]\((x + 2y)^{10}\)[/tex], we need to find [tex]\(T_{k+1}\)[/tex] where [tex]\(k+1 = 7\)[/tex]. This implies [tex]\(k = 6\)[/tex].
The general term [tex]\(T_{k+1}\)[/tex] in the binomial expansion is given by:
[tex]\[
T_{k+1} = \binom{n}{k} x^{n-k} (2y)^k
\][/tex]
Substitute [tex]\(n = 10\)[/tex] and [tex]\(k = 6\)[/tex]:
[tex]\[
T_7 = \binom{10}{6} x^{10-6} (2y)^6
\][/tex]
Calculate each part of the term:
1. The binomial coefficient [tex]\(\binom{10}{6}\)[/tex]:
[tex]\[
\binom{10}{6} = \frac{10!}{6!(10-6)!} = \frac{10!}{6!4!}
\][/tex]
Using precomputed factorial values, we find:
[tex]\[
\binom{10}{6} = 210
\][/tex]
2. The power of [tex]\(x\)[/tex]:
[tex]\[
x^{10-6} = x^4
\][/tex]
3. The power of [tex]\(2y\)[/tex]:
[tex]\[
(2y)^6 = 2^6 y^6 = 64y^6
\][/tex]
Combine all parts together:
[tex]\[
T_7 = 210 x^4 64 y^6
\][/tex]
The coefficient of the seventh term is the product of the binomial coefficient and the constant from [tex]\((2y)^6\)[/tex]:
[tex]\[
\text{Coefficient} = 210 \times 64 = 860160
\][/tex]
Thus, the coefficient of the seventh term in the expansion of [tex]\((x + 2y)^{10}\)[/tex] is [tex]\(860160\)[/tex].
However, provided options do not have a coefficient of 860160, indicating 'e. none' is correct answer.
