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Find the value of $$\sum_{i=0}^{24}\binom{200}{4i+2}$$

Actually there was a problem that I was solving and it has a sub part. It was to find the value of $\sum_{r=1}^{25}\binom{200}{8r-6}$

After a lot of rigorous dedication, I transformed the above expression into the original problem. I could not proceed further.

Any help is greatly appreciated.

MathStackexchangeIsNotSoBad
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  • The values in the sum increase more than exponentially for subsequent iterations; ${{200}\choose{8r - 6}} = \frac{200!}{(206 - 8r)! \cdot (8r - 6)!}$. $r = 1$ outputs $19900$ and $r = 2$ outputs $22451004309013280$. – Dstarred Jan 20 '23 at 14:02
  • I apparently have some answer it is like $2^{197}-2^{98}$ smt.. @Dstarred – MathStackexchangeIsNotSoBad Jan 20 '23 at 14:04
  • In powers of $2$ hmmm ..... here is fact; ${{n}\choose{k}} = 2^{p}, p, n, k \in \mathbb{N}$ only when $k = 1$ or $k = n - 1$. I believe there are multiple combinations of this in the sum, henceforth the large power. – Dstarred Jan 20 '23 at 14:19
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    Try to expand $(1+1)^{200}$, $(1-1)^{200}$, $(1+i)^{200}$ and $(1-i)^{200}$. I think this is the way, but I didn't do the calculations. – Sávio Jan 20 '23 at 14:31

2 Answers2

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We will use $i=\sqrt{-1}$. Then define:

$$A=(1+1)^{200}=\sum_{k=0}^{200}{200 \choose k} 1^k$$

$$B=(1-1)^{200}=\sum_{k=0}^{200}{200 \choose k} (-1)^k$$

$$C=(1+i)^{200}=\sum_{k=0}^{200}{200 \choose k} i^k$$

$$D=(1-i)^{200}=\sum_{k=0}^{200}{200 \choose k} (-i)^k$$

Now take A+B-C-D:

$$2^{200} + 0-(1+i)^{200}-(1-i)^{200}=\sum_{k=0}^{200}{200 \choose k} (1+(-1)^k-i^k-(-i)^k)$$

Note that $(1+(-1)^k-i^k-(-i)^k)$ is equal to zero unless $k \equiv 2 \pmod 4$, when it is $4$.

$$2^{200} -2 \times 2^{100}=8\sum_{j=0}^{24}{200 \choose {4j+2}} $$

$$\sum_{j=0}^{24}{200 \choose {4j+2}} =2^{197} - 2^{98} $$

Blitzer
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Consider the polynomial $f(x) = x^2(1+x)^{200}$. Notice that for $i \geq 1$ the coefficient $a_{4i}$ of $x^{4i}$ is $\binom{200}{4(i-1)+2}$. So we would like to get the sum of the coefficients $a_4, a_8, \dots, a_{100}$. Also note that due to the symmetry $$ \binom{n}{k} = \binom{n}{n-k} $$ we have $$ S = \sum_{i = 0}^{24} \binom{200}{4i+2} = \sum_{i=25}^{49} \binom{200}{4i+2} $$ and so $$\sum_{i=1}^{25} a_{4i} = \sum_{i=26}^{50} a_{4i}$$

Therefore $$ S = \frac{1}{2}\sum_{i=1}^{\infty} a_{4i}$$ because bigger coefficients are $0$.

Now let us consider the expression $$L = \frac{f(1) + f(\omega) + f(\omega^2) + f(\omega^3)}{4},$$ where $\omega = \exp{(\frac{2i\pi}{4})}$ is the fourth root of unity (here $i$ is the imaginary unit). Note that $S = \frac{L}{2}$, based on e.g. this. So we calculate $$ L = \frac{2^{200} + \omega^2(1+\omega)^{200} + (1+\omega^2)^{200} + \omega^2 (1+\omega^3)^{200}}{4} $$ Note that $\omega^2 = -1$ and $\omega^2 (1+\omega)^{200} = \omega^2 (1+\omega^3)^{200} = -2^{100}$. So the result is $$ \frac{L}{2} = \frac{2^{200} - 2\cdot 2^{100}}{8} = 2^{197} - 2^{98} $$

NikoWielopolski
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  • This method is called the roots of unity filter. The offset can be accounted for by introducing a factor in $f(x)$ – DatBoi Jan 20 '23 at 15:20