Is this a correct way to rewrite the binomial theorem?

AI Thread Summary
The discussion focuses on the correct rewriting of the binomial theorem, specifically extracting the first term from the summation. The proposed rewrite, (a+b)^n = {n choose 0}a^n + ∑_{k=1}^{n}{n choose k} a^{n-k} b^k, is confirmed as valid. Participants also verify another expression involving sigma notation, concluding that both transformations are correct. The conversation highlights the importance of understanding sigma notation in manipulating binomial expansions. Overall, the participants successfully clarify their queries regarding the binomial theorem.
jey1234
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Homework Statement


I am doing a poof and I need to use the binomial theorem. However is the following a correct way to rewrite it?

(a+b)^n\ =\ {n \choose 0}a^{n} + \sum_{k=1}^{n}{n \choose k}\ a^{n-k}\ b^{k}

Homework Equations



(a+b)^n\ =\ \sum_{k=0}^{n}{n \choose k}\ a^{n-k}\ b^{k}

The Attempt at a Solution


Basically, I want to extract the first term of the binomial expansion out of the summation but I'm not that good with sigma notation. Don't I just have to increase the lower bound by 1 and write the first term outside (as shown above)? Thanks.
 
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hey jey1234! :wink:

yes, that's fine :smile:
 
thanks tim. one more quick question. is the following correct?

\frac{1}{a}\sum_{k=1}^{n}a^{k}\ b^{k}\ =\ \sum_{k=1}^{n}a^{k-1}\ b^{k}
 
jey1234 said:
thanks tim. one more quick question. is the following correct?

\frac{1}{a}\sum_{k=1}^{n}a^{k}\ b^{k}\ =\ \sum_{k=1}^{n}a^{k-1}\ b^{k}
Yes, that's correct.
 
thanks sammy :smile:
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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