# problem of binomial coefficients

Find the value of $$\sum_{r=0}^5\binom{32}{6r}$$

Now I know this is a very easy question, but unfortunately I'm not getting the right answer.

I considered expansions of $$(1+1)^{32},(1+\omega)^{32},(1+\omega^2)^{32},(1+\omega^3)^{32},(1+\omega^4)^{32},(1+\omega^5)^{32}$$

After adding all of the above mentioned terms and dividing the result by $$6$$ should get us the required answer.

But when I am performing the above mentioned method, I am getting $$\frac{2^{33}-2}{6}$$ as answer which does not matches that of wolframaplha.

Any help is greatly appreciated.

• Er… what is $\omega$? Feb 5, 2023 at 18:22
• @jpboucheron cube root of unity Feb 5, 2023 at 18:23
• Indeed, in SageMath, command numerical_approx(sum((1+exp(2*i*k*pi/3))^32 , k , 0 , 5), digits=15) provides 8589934590; as for sum(binomial(32,6*k) , k, 0, 5) it gives 708653429. Feb 5, 2023 at 18:27
• Your idea was fine with $\omega=\exp\big(i\frac{2\pi}{3}\big)$, sixth root of unity, but you miscalculated the sum $\sum_{n=0}^5(1+\omega^n)^{32}$. Feb 5, 2023 at 19:11
• @jpboucheron pls tell me the correct method Feb 5, 2023 at 19:17

Let $$\omega=\exp\big(i\frac{\pi}3\big)\cdot$$ Observe that, for any integer $$k$$ we have $$\sum_{n=0}^5 \omega^{nk}= \begin{cases} \frac{\omega^{6k-1}}{\omega^k-1}=0&\text{if}\ k\notin6\mathbb{Z},\\[3pt] 6&\text{if}\ k\in6\mathbb{Z}. \end{cases}$$

From the Newton binomial formula $$\begin{eqnarray*} \sum_{n=0}^5(1+\omega^n)^{32} &=&\sum_{n=0}^5\sum_{k=0}^{32}\binom{32}{k}\omega^{nk} =\sum_{k=0}^{32}\binom{32}{k}\sum_{n=0}^5\omega^{nk}\\ &=&\kern-0.7em\sum\limits_{k\in\{0,6,12,18,24,30\}}\kern-0.6em \binom{32}{k}\times6\\ &=&6\sum_{r=0}^5\binom{32}{6r}. \end{eqnarray*}$$

So the required sum is $$S=\big(\sum_{n=0}^5(1+\omega^n)^{32}\big)\big/6$$.

Now we have $$1+\omega^s =\left\{ \begin{array}{rcl} 2&\text{if}\ s=0,\\[3pt] \sqrt3\exp\big(i\frac\pi6\big)&\text{if}\ s=1,\\[3pt] \exp\big(i\frac\pi3\big)&\text{if}\ s=2,\\[3pt] 0&\text{if}\ s=3,\\[3pt] \exp\big({-}i\frac\pi3\big)&\text{if}\ s=4,\\[3pt] \sqrt3\exp\big({-}i\frac\pi6\big)&\text{if}\ s=5. \end{array} \right.$$

In the end we obtain $$\begin{eqnarray*} S&{}={}&\bigg[2^{32} +3^{16}\!\exp\Big({-}i\frac{2\pi}3\Big) +\exp\Big(i\frac{2\pi}3\Big) +0 +\exp\Big({-}i\frac{2\pi}3\Big) +3^{16}\!\exp\Big(i\frac{2\pi}3\Big)\bigg]\mathbin{\Big/}6\\ &{}={}&\frac{2^{32}-3^{16}-1}6\\ &{}={}&708653429. \end{eqnarray*}$$

Edit (to answer a comment below).

I don't see any valuable generalization of the method above. Suppose that one replaces 32 by $$m$$, 6 by $$\lambda$$ and puts $$\ell=\lfloor m\lambda\rfloor$$, in order to calculate $$\sum_{n=0}^\ell\binom{m}{\lambda n}$$. We would take $$\omega=\exp(2i\pi/\lambda)$$, but the complication is that there's no reason in general to have $$\sum\limits_{n\leqslant\ell}\omega^{kn}=0$$ for $$k\notin\lambda\mathbb{Z}$$, as above. In general this sum would be $$\frac{\omega^{k(\ell+1)}-1}{\omega^k-1}\cdot$$

With the values $$(m,\lambda)=(32,6)$$ given in the problem, everything went fine because $$\ell+1\in\lambda\mathbb{Z}$$ (i. e. $$5+1\in6\mathbb{Z}$$.

Perhaps the general $$\displaystyle\sum_{n=0}^\ell\binom{m}{\lambda n}$$ can be calculated by other means?

• I see that you've here considered 6th root of unity...can you pls add a generalization...i implore you Feb 6, 2023 at 11:46

Having only 6 terms, I think the most straightforward way is to look at each coefficient individually. Let's call $$S$$ the desired sum.

$$S = \sum_{r=0}^5 \binom{32}{6r} = \binom{32}{0} + \binom{32}{6} + \binom{32}{12} + \binom{32}{18} + \binom{32}{24} + \binom{32}{30}$$

This shows that:

$$S = 1 + \dfrac{32!}{26!\cdot6!} + \dfrac{32!}{20!\cdot12!} + \dfrac{32!}{14!\cdot18!} + \dfrac{32!}{8!\cdot24!} + \dfrac{32!}{2!\cdot30!}$$

If we take $$\dfrac{32!}{26!\cdot6!}$$, it equals $$\dfrac{32\cdot31\cdot30\cdot29\cdot28\cdot27}{6\cdot5\cdot4\cdot3\cdot2\cdot1} = 906192$$.

Repeating the same process, we get $$\binom{32}{12} = 225792840$$, $$\binom{32}{18} = 471435600$$, $$\binom{32}{24} = 10518300$$, $$\binom{32}{30} = \dfrac{32\cdot31}{2} = 16\cdot 31 = 496$$.

Finally, $$\sum_{r=0}^5 \binom{32}{6r} = 1+906192+225792840+471435600+10518300+496 = 708653429$$

But all of this should not be done by hand, and has not been, it should instead be done with a computer, and seeing the prime factorization of the result, I really don't see how to bypass the steps mentioned above.