A probability problem by Chan Tin Ping

Probability Level 3

A = ∑ k = 100 ∞ ( k 100 ) ( 1 2 ) k − 100 \large A=\sum_{k=100}^{\infty}\binom{k}{100}\left(\frac{1}{2}\right)^{k-100}

Find log ⁡ 2 A \log_2A .


The answer is 101.

Chan Tin Ping by Chan Tin Ping , 20, Malaysia

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1 solution

Mark Hennings
Sep 23, 2017

Since, for any positive integer n n , ∑ k = n ∞ ( k n ) x k − n = ∑ k = 0 ∞ ( n + k n ) x k = ( 1 − x ) − n − 1 ∣ x ∣ < 1 \sum_{k=n}^\infty \binom{k}{n} x^{k-n} \; = \; \sum_{k=0}^\infty \binom{n+k}{n} x^k \; = \; (1 - x)^{-n-1} \hspace{2cm} |x| < 1 we deduce that A = 2 101 A = 2^{101} , making the answer 101 \boxed{101} .

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https://en.m.wikipedia.org/wiki/Binomial_theorem

Chan Tin Ping - 3 years, 8 months ago

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