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Find the 6th term in the expansion of
(x+2y)^(8) in simplest form.
Answer:

Find the $6$ th term in the expansion of $(x+2 y)^{8}$ 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

(x+2 y)^{8}

in simplest form.

\newline

Answer:

Use Binomial Theorem: To find the $6^{\text{th}}$ term in the expansion of $(x+2y)^{8}$ , we will use the binomial theorem. The binomial theorem states that the $n^{\text{th}}$ term in the expansion of $(a+b)^{m}$ is given by the formula $C(m, k) \cdot a^{m-k} \cdot b^k$ , where $C(m, k)$ is the binomial coefficient ＂ $m$ choose $k$ ＂. For the $6^{\text{th}}$ term, $k$ will be $(x+2y)^{8}$ $0$ because the first term corresponds to $(x+2y)^{8}$ $1$ .
Calculate Binomial Coefficient: First, we need to calculate the binomial coefficient $C(8, 5)$ . This is equal to $\frac{8!}{5! \times (8-5)!}$ , which simplifies to $\frac{8!}{5! \times 3!}$ .
Calculate Factorials: Calculating the factorial values, we get $8! = 8 \times 7 \times 6 \times 5!$ , $5! = 5 \times 4 \times 3 \times 2 \times 1$ , and $3! = 3 \times 2 \times 1$ . Substituting these into the binomial coefficient formula, we have $C(8, 5) = \frac{8 \times 7 \times 6}{3 \times 2 \times 1}$ .
Calculate Binomial Coefficient: Simplifying the fraction, we get $C(8, 5) = \frac{8 \times 7 \times 6}{3 \times 2} = 56$ .
Calculate Remaining Term: Now we need to calculate the rest of the term, which is $x^{(8-5)} \times (2y)^5$ . This simplifies to $x^3 \times (2y)^5$ .
Calculate Power of $2$ y: Raising $2y$ to the $5$ th power gives us $2^5 \times y^5$ , which is $32 \times y^5$ .
Multiply Coefficients: Multiplying the binomial coefficient by the powers of $x$ and $y$ , we get the $6$ th term: $56 \times x^3 \times 32 \times y^5$ .
Calculate Final Term: Finally, we multiply $56$ by $32$ to get the coefficient of the $6$ th term. This gives us $56 \times 32 = 1792$ .
Final Simplified Term: The $6^{th}$ term in the expansion of $(x+2y)^{8}$ in simplest form is therefore $1792 \times x^3 \times y^5$ .

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