I am a high school student and I am searching for a not very complicated way to calculate $${100\choose 0}^2-{100\choose 1}^2+{100\choose 2}^2-{100\choose 3}^2+...+{100\choose 100}^2.$$
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2this question gives you a way to proceed. – lulu Mar 11 '20 at 16:43
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I searching for a simpler answer but thanks. – tantan Mar 11 '20 at 16:44
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http://www.wolframalpha.com/input/?i=sum%20from%20i%3D0%20to%20k%20%28%28-1%29%5Ei%20binomial%28k%2Ci%29%5E2%29%20and%20k%3D100 – sampleuser Mar 11 '20 at 16:47
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1That's a really simple method though... I don't know of a straight combinatorial interpretation of the result (though of course there might be one). – lulu Mar 11 '20 at 16:47
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What specifically do you not understand about the answers in the linked question? For what reason do you reject those answers and want a "simpler" approach? How much "simpler" do you expect the answer to be? $\binom{100}{50}$ is already simplified greatly. – JMoravitz Mar 11 '20 at 16:59
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3As an aside, this is the second time you asked this within the past hour. You should not just repost the same question without addressing the issues that it was previously closed for. Repeated violation of this is inviting consequences. – JMoravitz Mar 11 '20 at 17:03
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If the explanations on that answer are still unclear, then try going through the linked questions for additional explanations such as this one or this one or others. – JMoravitz Mar 11 '20 at 17:08
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Let $$S_n=\sum_{k=0}^{n} (-1)^k {n \choose k}^2$$ $$(1+x)^n=\sum_{k=0}^{n} {n \choose k} x^{k}~~~~~)1)$$ $$(1-1/x)^n=\sum_{k=0}^{n} (-1)^k {n \choose k} x^{-k}~~~~~)1)$$ Multiplying (1) and (2), we get $$(-1)^n (1-x^2)^n= \sum_{k=0}^{n} (-1)^k {n \choose k}^2 x^0+...+...$$ So $$S_n=[x^0] (-1)^n (1-x^2)^n=(-1)^{n/2}{n \choose n/2},~ if~ n ~ is ~ even$$ So the sum of the required series is $${100 \choose 50}$$
Z Ahmed
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