I came across this problem in which we had to evaluate $$\sum\limits_{k=3}^{k=7} {(k+1)\choose (7-k)}$$ All I could think of was to put in the values of k and get $${4\choose4}+{5\choose3}+{6\choose2}+{7\choose1}+{8\choose0} = 34$$ Is there a more efficient way (possibly using binomial concept) to solve this sum?
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There is no better way. – Kavi Rama Murthy Dec 23 '19 at 07:24
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1Fibonnacci may help you – MafPrivate Dec 23 '19 at 07:27
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2@IsaacYIUMathStudio Could you elaborate pls? – HSB Dec 23 '19 at 07:30
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1This is $\sum\limits_{k=\lfloor n/2\rfloor}^{k=n} {k+1\choose n-k}$ when $n=7$. Try it for other values of $n$ and you may understand @IsaacYIUMathStudio's comment – Henry Dec 23 '19 at 07:32
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So does it mean that every sum of this form that @henry has given in his comment can be evaluated using Fibonacci? – HSB Dec 23 '19 at 07:44
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1I would simply use a computer, just as we often use computers for other large sums. – David G. Stork Dec 23 '19 at 08:00