The question says to find the value of $$\binom{n}{1} \cdot \left( \sum ^1 _ {r=0} \binom{1}{r}\right) + \binom{n}{2} \cdot \left( \sum ^2 _ {r=0} \binom{2}{r}\right) + \binom{n}{3} \cdot \left( \sum ^3 _ {r=0} \binom{3}{r}\right) \ldots \binom{n}{n} \cdot \left( \sum ^n _ {r=0} \binom{n}{r}\right)$$
I wrote $$\left( \sum ^n _ {r=0} \binom{n}{r}\right) = 2^n$$ Which gives us a series like this,
$$\binom{n}{1} \cdot 2 + \binom{n}{2} \cdot 2^2 + \ldots \binom{n}{n} \cdot 2^n$$
But I am not able to figure out what to do from here.
Any help would be much appreciated.
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Prakhar Nagpal
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How would you expand $(1+2)^n?$ using the Binomial Theorem? – Oct 02 '18 at 17:09
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@AnotherJohnDoe Using the Binomial expansion I expect I would get something, like, $\binom{n}{0} \cdot 1 + \binom{n}{1} \cdot 2 + \ldots + \binom{n}{n-1} \cdot 2^{n-1} + \binom{n}{n} \cdot 2^n$ – Prakhar Nagpal Oct 02 '18 at 17:11
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I understand. Thank you. – Prakhar Nagpal Oct 02 '18 at 17:13
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Mathematica yields $3^n-1$. – David G. Stork Oct 02 '18 at 17:13
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Obviously since the first term is missing. Thank you – Prakhar Nagpal Oct 02 '18 at 17:14