Two large numbers

Algebra Level 4

Which of these numbers is bigger ? ? \\ ( A ) 2 9999 OR ( B ) 3 ∑ n = 0 3333 ( 9999 3 n ) (A) \quad \quad \large 2^{9999} \\ \text{OR} \\ (B) \quad \quad \large 3\displaystyle\sum^{3333}_{n=0}\dbinom{9999}{3n}

The are the same ( A ) (A) is larger ( B ) (B) is larger

Akeel Howell by Akeel Howell , 22, USA

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1 solution

Akeel Howell
Mar 5, 2017

3 ∑ n = 0 3333 ( 9999 3 n ) = 3 ( 1 + 1 ) 9999 + ( 1 + ω ) 9999 + ( 1 + ω 2 ) 9999 3 = 3 2 9999 + ( − ω 2 ) 9999 + ( − ω ) 9999 3 = 3 ( 2 9999 − 1 − 1 ) 3 = 2 9999 − 2 2 9999 − 2 < 2 9999 \\ \large\displaystyle 3\sum^{3333}_{n=0}\dbinom{9999}{3n} = 3\dfrac{(1+1)^{9999}+(1+\omega)^{9999}+(1+\omega^2)^{9999}}{3} \\ = 3\dfrac{2^{9999}+(-\omega^2)^{9999}+(-\omega)^{9999}}{3} =3 \dfrac{(2^{9999}-1-1)}{3} = 2^{9999}-2 \\ 2^{9999}-2 < 2^{9999}

ω \omega is a third root of unity.

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