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

Find the $2$ nd term in the expansion of $(2 x-5 y)^{5}$ 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

(2 x-5 y)^{5}

in simplest form.

\newline

Answer:

Use Binomial Theorem: To find the $2$ nd term in the expansion of $(2x-5y)^{5}$ , 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: First, 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.
Calculate Powers of a and b: Calculating $C(5, 1)$ gives us $\frac{5!}{1! * (5-1)!} = \frac{5}{1 * 4!} = \frac{5}{1 * 24} = \frac{5}{24}$ . Since $24$ is the factorial of $4$ , we simplify it to get $\frac{5}{24} = 5$ .
Multiply Coefficient and Powers: Now we need to calculate the powers of $a$ and $b$ for the $2$ nd term. Since $a = 2x$ and $b = -5y$ , and we are looking for the $2$ nd term where $k=2$ , we have $a^{5-2+1} = a^{4}$ and $b^{2-1} = b^{1}$ .
Simplify the Expression: Calculating the powers, we get $(2x)^{4} = 16x^{4}$ and $(-5y)^{1} = -5y$ .
Simplify the Expression: Calculating the powers, we get $(2x)^{4} = 16x^4$ and $(-5y)^{1} = -5y$ .Multiplying the binomial coefficient by the powers of $a$ and $b$ , we get the $2$ nd term: $T(2) = C(5, 1) \cdot (2x)^{4} \cdot (-5y)^{1} = 5 \cdot 16x^4 \cdot -5y$ .
Simplify the Expression: Calculating the powers, we get $(2x)^{4} = 16x^4$ and $(-5y)^{1} = -5y$ .Multiplying the binomial coefficient by the powers of $a$ and $b$ , we get the $2$ nd term: $T(2) = C(5, 1) \cdot (2x)^{4} \cdot (-5y)^{1} = 5 \cdot 16x^4 \cdot -5y$ .Simplifying the expression, we get $T(2) = 5 \cdot 16 \cdot -5 \cdot x^4 \cdot y = -400 \cdot x^4 \cdot y$ .

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