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Mathematics 13 Online
OpenStudy (anonymous):

The first question asked me to approximate 2nCn using stirling's formula and i got 1/sqrt(pi*n). The next question asked me to approximate (2nCn)^2. Can I use my previous answer and square it, i.e., (1/sqrt(pi*n))^2 or do I have to redo Stirling's formula again? Can i do the same for 4nC2n (multiplied by 2)?

OpenStudy (anonymous):

I did the work, figured out the answer is yes to anyone wondering

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