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Write $\sum\limits_{k=0}^{1000} \binom{1000}{k}5^k$ as $A^B$, where have to find $A$ and $B$.

I would love to tell you what I've tried, but I don't even understand the question, so have no idea where to begin. I know I could write $\sum\limits_{k=0}^{1000} \binom{1000}{k}$ as $2^{1000}$, but I don't see how that helps. Feeling pretty stupid :(

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    $\begingroup$ Just take $A = \sum_{k=0}^{1000} \binom{1000}{k}5^k$ and $B = 1$. $\endgroup$
    – Mees de Vries
    Commented Oct 5, 2017 at 9:00
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    $\begingroup$ All silliness aside, are you familiar with the binomial theorem? $\endgroup$
    – Mees de Vries
    Commented Oct 5, 2017 at 9:01
  • $\begingroup$ @MeesdeVries Yes, that is what we've been studying $\endgroup$
    – XRBtoTheMOON
    Commented Oct 5, 2017 at 9:02
  • $\begingroup$ Does the expression you've been given look like one side of the equation in the binomial theorem? That would be a good place to start. $\endgroup$
    – Mees de Vries
    Commented Oct 5, 2017 at 9:03
  • $\begingroup$ @MeesdeVries I don't understand this $A$ and $B$ thing tho? How can I express that as some sort of variable we have to find? $\endgroup$
    – XRBtoTheMOON
    Commented Oct 5, 2017 at 9:09

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Look at $(1+5)^{1000}$. Binomial theorem !

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  • $\begingroup$ Thank you very much. I see it now. $\endgroup$
    – XRBtoTheMOON
    Commented Oct 5, 2017 at 9:18

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