Binomial Theorem of (3n + 2)^x

Natasha1
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Could anyone tell me what (3n + 2)^x equals to please so I can check my answer?

I get something awful that would take me too long to type
 
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x is a positive integer right ?

Well just apply THIS

marlon
 
marlon said:
x is a positive integer right ?

Well just apply THIS

marlon

oops sorry forgot to say x is an integer therefore +ve or -ve. I know the formula just wanted to triple check with a math expert to see if I hadn't made a mistake that's all.
 
Natasha1 said:
oops sorry forgot to say x is an integer therefore +ve or -ve. I know the formula just wanted to triple check with a math expert to see if I hadn't made a mistake that's all.

Then you phrased your question very badly. If you want to check your answer, tell us what your answer is. We tend to get very suspicious of someone asking us to tell the answer so he can "check" his answer!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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