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The binomial theorem states that

$$(A+B)^n=\sum_{k=1}^{n}{n \choose k}A^{n-k}B^k$$

I need help expressing the following summation as a modified binomial expression:

$$\sum_{k=1}^{n}n!{n-1\choose k}A^{k+1}B^{n+k}=(A?B)^?$$


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You need "$k=0$" rather than "$k=1$". – Michael Hardy Jan 4 '12 at 0:43
up vote 4 down vote accepted

Your first expression should be $(A+B)^n=\sum_{k=0}^{n}{n \choose k}A^{n-k}B^k$ starting from $k=0$.

Your second expression may have problems with $k=n$ when evaluating ${n-1 \choose n}$.

So I will try to help with a slightly altered version of your question: $$\sum_{k=0}^{n-1}n!{n-1\choose k}A^{k+1}B^{n+k} = n!A B^n \sum_{k=0}^{n-1}{n-1\choose k}1^{n-1-k}(AB)^{k} = n!A B^n (1+AB)^{n-1}.$$

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Thanks! But what is the $1^{n-1-k}$ term doing? Doesn't that just always equal 1, for all n and k? – ben Jan 4 '12 at 0:56
It is there to make it obvious we are applying the earlier binomial expansion of $(1+AB)^{n-1}$ . It can be ignored if necessary – Henry Jan 4 '12 at 8:30

$$ \sum_{k=1}^n n! \binom{n-1}{k} A^{k+1} B^{n+k} = n! \cdot A \cdot B^n \cdot \sum_{k=1}^{n-1} \binom{n-1}{k} A^{k} B^{k} = n! \cdot A \cdot B^{n} \left( (1+A B)^{n-1} - 1 \right) $$

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