I know that (n+1)! = n!(n+1) but how would i solve the factorial: (2n+3)! thank you.

I don't know what you mean by 'solve' in this case.

(2n+3)! = (2n)!(2n+1)(2n+2)(2n+3) if that helps.

thx man

To solve the factorial (2n + 3)!, we can write it as a multiplication of consecutive numbers starting with 1 and ending with (2n + 3). Here is how you can calculate it step by step:

1. Start with 1 and multiply it by the next number, which is 2. So far, the product is 1 * 2 = 2.

2. Multiply the previous result by the next number, which is 3. The current product is 2 * 3 = 6.

3. Continue this process, multiplying the current product by the next consecutive number all the way up to (2n + 3).

4. Repeat step 3 until you reach (2n + 3). Each step will look like (previous product) * (current number).

5. Finally, you will have the value of (2n + 3)!.

For example, If you want to find the value of (2n + 3)! when n = 2:
- Start with 1 and multiply it by 2, then multiply the result by 3, resulting in 1 * 2 * 3 = 6.
- Next, multiply the current result, 6, by 4, then multiply the new result by 5. The product becomes 6 * 4 * 5 = 120.
- Finally, multiply the current result, 120, by 6, then multiply the new result by 7. The final product is 120 * 6 * 7 = 5040.

So, when n = 2, (2n + 3)! equals 5040.