How Do You Solve Advanced Binomial Expansion Problems?

AI Thread Summary
To solve advanced binomial expansion problems, one can use the binomial theorem, which provides a formula for expanding expressions like (a + b)^n. For (3x + 4)^4, the first four terms in descending order can be derived by applying the binomial coefficients and the powers of x accordingly. When asked to find expansions in descending order, it is not necessary to calculate all terms up to n; instead, focus on the required number of terms by using the coefficients in reverse. The binomial coefficients are symmetric, meaning they mirror each other, which simplifies calculations. Understanding the general formula and applying combinatorial methods for coefficients is key to tackling these problems effectively.
saltrock
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Binomial Expansion problems!Please help me ASAP!

1)Using binomial expand (3X +4)^4.can you please kindly explain your process.
2)Use the first 4 terms in the binomial expansion of (3x+4)^4 in the DESCENDING order of x to determine the approx. of 1.004^12.Can you please give me the way how to tackle a problem with descending powe.I know how to do ascending.IF it says first 3 terms in the ascending order i can easily get it by using the formula but if it is like ,'find the expansion of(>>blah blah)^20 in descending order,does this mean that we have to find the values all the way to 20 and get the last 3 terms or is there any easier way to do it?
3)EXpand using binomial expansion
( root3 - 2)^6

I'd like to thank in advance if anyone can help me do this.cheers.
 
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The really neat thing about the binomial expansion is that the binomial coefficients are the same in reverse order as in the forward direction, i.e. the first and last are equal, the second and next to last are equal and so on.
 
Tide is right here. Let me add this : if you want to calculate a binomial expansion just look at the general formule and try to "fit" the given data into this formula. Then just apply the definition (the expansion) and the only thing you need to calculate are the coefficients using the combinatorics...

regards
marlon
 
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