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Find the 4 th term in the expansion of
(x-3)^(8) in simplest form.
Answer:

Find the $4$ th term in the expansion of $(x-3)^{8}$ in simplest form. $\newline$ Answer:

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Partial sums of geometric series

Full solution

Q. Find the

4

th term in the expansion of

(x-3)^{8}

in simplest form.

\newline

Answer:

Use Binomial Theorem: To find the $4$ th term in the expansion of $(x-3)^{8}$ , we will use the binomial theorem. The binomial theorem states that the $k$ th term in the expansion of $(a+b)^{n}$ is given by $C(n, k-1) \cdot a^{(n-k+1)} \cdot b^{(k-1)}$ , where $C(n, k)$ is the binomial coefficient ＂ $n$ choose $k$ ＂. For the $4$ th term, $k = 4$ .
Calculate Binomial Coefficient: First, we calculate the binomial coefficient $C(8, 4-1)$ which is $C(8, 3)$ . This is equal to $\frac{8!}{3! \times (8-3)!}$ , where $\text{“!“}$ denotes factorial.
Simplify Factorials: Calculating the factorial values, we get $8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$ , $3! = 3 \times 2 \times 1$ , and $(8-3)! = 5! = 5 \times 4 \times 3 \times 2 \times 1$ .
Calculate Binomial Coefficient: Now, we simplify the binomial coefficient $C(8, 3) = \frac{8!}{3! \times 5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$ .
Calculate Powers of a and b: Next, we calculate the remaining parts of the term: $a^{(n-k+1)} = x^{(8-4+1)} = x^5$ and $b^{(k-1)} = (-3)^{(4-1)} = (-3)^3$ .
Simplify Powers: Now, we simplify $(-3)^3 = -27$ .
Multiply Coefficients: Finally, we multiply the binomial coefficient by the powers of $a$ and $b$ to get the $4$ th term: $56 \times x^5 \times (-27)$ .
Final Result: Multiplying $56$ by $-27$ , we get $-1512$ . So, the $4$ th term in the expansion is $-1512 \times x^5$ .

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