It is quite well-known that
$(x_1 + \ldots + x_n)^2 = x_1^2 + \ldots + x_n^2 + 2 \sum_{i=1}^{n}\sum_{j=1}^{i}x_ix_j$.
Is there a similarly elegant way to express $(x_1 + \ldots +x_n)^4$?
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1Not as elegant, but the multinomial theorem does give an expression for this, yes. – Rushabh Mehta Feb 09 '22 at 21:56
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1Just apply the previous identity twice, even though it's not thaaat nice. – Manatee Pink Feb 09 '22 at 22:06
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1You want the second sum to be $j=1$ to $i-1$, not $i$. – Robert Israel Feb 09 '22 at 22:13
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The formula that you showed was derived by noticing that $(a+b)^2 = a^2 + 2ab + b^2.$ The other formula may be similarly derived by noticing that $(a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4.$ – user2661923 Feb 09 '22 at 22:57
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https://math.stackexchange.com/a/3532682/995003 – mio Feb 10 '22 at 03:57