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Find the 6 th term in the expansion of
(4x-1)^(7) in simplest form.
Answer:

Find the $6$ th term in the expansion of $(4 x-1)^{7}$ in simplest form. $\newline$ Answer:

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Partial sums of geometric series

Full solution

Q. Find the

6

th term in the expansion of

(4 x-1)^{7}

in simplest form.

\newline

Answer:

Use Binomial Theorem: To find the $6^{\text{th}}$ term in the expansion of $(4x-1)^{7}$ , we will use the binomial theorem. The general term in the expansion of $(a+b)^{n}$ is given by $T(k+1) = \binom{n}{k} \cdot a^{n-k} \cdot b^{k}$ , where $\binom{n}{k}$ is the binomial coefficient ＂n choose k＂. For the $6^{\text{th}}$ term, $k = 5$ , since we start counting from $k = 0$ for the first term.
Calculate Binomial Coefficient: First, we calculate the binomial coefficient for $n = 7$ and $k = 5$ , which is $7C5$ . This can be calculated as $\frac{7!}{5! \times (7-5)!}$ .
Find $7C5$ : Calculating $7C5$ , we get $\frac{7!}{5! \cdot 2!} = \frac{7 \cdot 6}{2 \cdot 1} = 21$ .
Calculate $6$ th Term: Now, we will use the binomial coefficient to find the $6$ th term. The $6$ th term is $T(6) = \binom{7}{5} \cdot (4x)^{7-5} \cdot (-1)^5$ .
Substitute Values: Substitute the values into the formula to get $T(6) = 21 \times (4x)^2 \times (-1)^5$ .
Simplify Term: Simplify the term: $T(6) = 21 \times 16x^2 \times (-1)$ .
Final Result: Multiplying the numbers together, we get $T(6) = -336x^2$ .

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