Find the 6th term of the binomial theorem (x+2)10

The binomial theorem states that the $(r+1)$-th term of $(a+b)^n$ is given by the formula $\binom{n}{r} a^{n-r} b^r$.
In this problem, we have $(x+2)^{10}$:

• $a = x$
• $b = 2$
• $n = 10$ We need to find the 6th term, so $r+1 = 6$, which means $r = 5$.

The 6th term is:

$$T_{6} = \binom{10}{5} x^{10-5} (2)^5$$

First, calculate $\binom{10}{5}$:

$$\binom{10}{5} = \frac{10!}{5!(10-5)!} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 2 \times 9 \times 2 \times 7 = 252$$

Next, calculate the powers:

$$x^{10-5} = x^5$$ $$(2)^5 = 32$$

$$T_{6} = 252 \cdot x^5 \cdot 32$$ $$T_{6} = 8064x^5$$

The 6th term of the binomial expansion $(x+2)^{10}$ is $\boxed{\mathbf{8064x^5}}$.