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Find the 2nd term in the expansion of
(x-6y)^(5) in simplest form.
Answer:

Find the $2$ nd term in the expansion of $(x-6 y)^{5}$ in simplest form. $\newline$ Answer:

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Partial sums of geometric series

Full solution

Q. Find the

2

nd term in the expansion of

(x-6 y)^{5}

in simplest form.

\newline

Answer:

Use Binomial Theorem: To find the $2$ nd term in the expansion of $(x-6y)^{5}$ , we will use the binomial theorem. The general form of the $k$ -th term in the expansion of $(a+b)^{n}$ is given by $T(k) = C(n, k-1) \cdot a^{(n-k+1)} \cdot b^{(k-1)}$ , where $C(n, k)$ is the binomial coefficient ＂ $n$ choose $k$ ＂. For the $2$ nd term, $k=2$ .
Calculate Binomial Coefficient: We calculate the binomial coefficient for the $2^{nd}$ term, which is $C(5, 2-1) = C(5, 1)$ . The binomial coefficient $C(n, k)$ is calculated as $\frac{n!}{k! \cdot (n-k)!}$ , where $\text{“!“}$ denotes factorial.
Simplify Coefficient: $C(5, 1)$ is calculated as $\frac{5!}{(1! * (5-1)!)} = \frac{5}{(1 * 4!)} = \frac{5}{(1 * 24)} = \frac{5}{24}$ . Since $24$ is not a factor of $5$ , we simplify this to $5$ .
Calculate Powers of a and b: Now we need to calculate the powers of $a$ and $b$ for the $2$ nd term. Since $a$ is $x$ and $b$ is $-6y$ , and we are looking for the $2$ nd term where $k=2$ , we have $a^{5-2+1} = x^{5-1} = x^4$ and $b^{2-1} = (-6y)^{1} = -6y$ .
Multiply Coefficient and Powers: Multiplying the binomial coefficient by the powers of $a$ and $b$ , we get the $2$ nd term: $T(2) = C(5, 1) \times x^4 \times (-6y) = 5 \times x^4 \times (-6y) = -30x^4y$ .

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