Find the value of $${100 \choose0}^2+{100 \choose2}^2+{100 \choose4}^2+\cdots+{100 \choose100}^2$$ I can calculate the sum of squares of all coefficients but not even coefficients.
Sum of squares of even combinatorial coefficients
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combinatorics
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1Hint: You want to know $\sum\limits_{k\in\left\{0,1,\ldots,n\right\} \text{ is even}} \dbinom{n}{k}^2$. It suffices to know $\sum\limits_{k=0}^{n} \dbinom{n}{k}^2$ and $\sum\limits_{k=0}^{n} \left(-1\right)^k \dbinom{n}{k}^2$. These are both known (e.g., Proposition 2.25 **(c)** and Corollary 7.10 in my *Notes on the combinatorial fundamentals of algebra*, https://github.com/darijgr/detnotes/releases/tag/2016-12-22 , but really any text on binomial coefficients). – 2017-01-01
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1The given expression is equal to the coefficient of $x^{100}$ in $(\frac{1}{4})[(x+1)^{100}+(x-1)^{100}][(1+x)^{100}+(1-x)^{100}]$ – 2017-01-01
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0@darijgrinberg Yeah. Thanks! That basically solved the question. – 2017-01-01
1 Answers
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The given expression is equal to the coefficient of $x^{100}$ in $$(\frac{1}{4})[(x+1)^{100}+(x-1)^{100}][(1+x)^{100}+(1-x)^{100}]$$
= coefficient of $ x^{100}$ in$ (\frac{1}{4})((1+x)^{100}+(1-x)^{100})^2$
$$=(\frac{1}{2}){200 \choose100} + (\frac{1}{2}){100 \choose50}$$