Alternate Odd Binomial Coefficient

Probability Level 4

$\binom{30}{1} - \binom{30}{3} + \binom{30}{5} - \cdots + \binom{30}{29} = \, ?$

Notation: $\dbinom MN = \dfrac {M!}{N! (M-N)!}$ denotes the binomial coefficient .

The answer is -32768.

1 solution

Chew-Seong Cheong
Oct 28, 2016

Consider the following:

$\begin{aligned} (1+x)^{30} & = {30 \choose 0} + {30 \choose 1}x + {30 \choose 2}x^2 + {30 \choose 3}x^3 + {30 \choose 4} + ... + {30 \choose 30} & \small {\color{#3D99F6}\text{Putting }x = i = \sqrt{-1}} \\ \implies (1+i)^{30} & = {30 \choose 0} + {30 \choose 1}i - {30 \choose 2} - {30 \choose 3}i + {30 \choose 4} + ... - {30 \choose 30} & \small {\color{#3D99F6}\text{Taking the imaginary part}} \\ \Im \left \{ (1+i)^{30} \right \} & = {30 \choose 1} - {30 \choose 3} + {30 \choose 5} + ... + {30 \choose 29} = S \end{aligned}$

$\begin{aligned} \implies S & = \Im \left \{ (1+i)^{30} \right \} = \Im \left \{ (2i)^{15} \right \} = \Im \left \{ 2^{15}(-i) \right \} = -2^{15} = \boxed{-32768} \end{aligned}$

The more standard approach which extends the "substitute suitable values of $x$ " is to consider $(1-i)^{30}$ .

This relates to the theory of characters on groups :)

Calvin Lin Staff - 4 years, 7 months ago

Alternate Odd Binomial Coefficient

The answer is -32768.

1 solution

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