Prove that $\sum_{i=0}^{63}f_{i}\cdot\left(n+i\right)^{5}=0$ 
For $n \geq 1$, let $f_m = (-1)^s$ where $s$ is the digital sum modulo $2$ of the binary representation of $m$. Prove that $$ \sum_{i=0}^{63}f_{i}\cdot\left(n+i\right)^{5}=0.$$

Since $s$ is the digital sum taken modulo $2$ we know that $s \in \{0,1\}$. I don't see a pattern in digital sum of the binary representation modulo $2$: $0,1,1,0,1,0,0,1,1,0,0,1,0,1,\ldots$. How do we prove the sum is equal to zero?
 A: Claim $1$:$X_0=\sum_{i=0}^{63}f_i=0$
Proof: The number of binary strings of length $6$  with digital sum even is $$\binom{6}{0}+\binom{6}{2}+\binom{6}{4}+\binom{6}{6}=32$$,while the number of digital string of length $6$ with odd digital sum is $$\binom{6}{1}+\binom{6}{3}+\binom{6}{5}=32.$$From here it immediately follows that $\sum_{i=0}^{63}f_i=0$.
Claim $2$:$X_1=\sum_{i=0}^{63}f_i\cdot i=0$
Proof: This sum is equal to $\sum_{k=0}^{5}2^k(a-b)$,where $$a=\text{number of strings of length}~5~\text{with odd digital sum}$$ and $$b=\text{number of strings of length}~5~\text{with even digital sum}.$$
(To obtain this just write each number $i$ in binary form and see how many times the number $2^k,0\le k\le 5$ appears with $+$ sign and $-$ sign.)
As in proof for claim $1$ it's easy to prove that $a=b$,so $a-b=0$.
Claim $3$:$X_2=\sum_{i=0}^{63}f_i\cdot i^2=0$
Proof: Note that the digital sum of $i$ and $i+32$ in binary have different parities for any $0\le i\le 31$(just note that the $32$ puts a $1$ in front of the binary representation of $i$).
Therefore $$\sum_{i=0}^{63}f_i\cdot i^2=\sum_{i=0}^{31}f_i\cdot i^2-\sum_{i=0}^{31}f_i(32+i)^2=-32^2\sum_{i=0}^{31}f_i-64\sum_{i=0}^{31}f_i\cdot i.$$
Now we just proceed as in proof of claim $1$ and proof of claim $2$ to prove that $\sum_{i=0}^{31}f_i=0$ and $\sum_{i=0}^{31}f_i\cdot i=0$.
Claim $4$: $X_3=\sum_{i=0}^{63}f_i\cdot i^3=0$
Proof: We do the same trick with $f_{32+i}=-f_i,\forall 0\le i\le 31$ as in proof of claim $3$.Then it reduces to proving claims $1,2,3$ for $2^5-1=31$ instead of $2^6-1=63$,but the proofs are almost identical.
Claim $5$: $X_4=\sum_{i=0}^{63}f_i\cdot i^4=0$
Proof: Same trick.
Claim $6$: $X_5=\sum_{i=0}^{63}f_i\cdot i^5=0$
Proof: Again,same trick.
Finally,from the above $6$ claims we have $$\sum_{i=0}^{63}f_i\cdot(n+i)^5=n^5X_0+5n^4X_1+10n^3X_2+10n^2X_3+5nX_4+X_5=0.$$
NOTE: This is a pretty much brute-force kind of method,but it uses only elementary stuff.The above proof can be easily modified to prove that $$\sum_{i=0}^{2^{k+1}-1}f_i\cdot (n+i)^{t}=0,\forall 0\le t\le k.$$
A: 
Lemma:  Let $r\geq 0$ be an integer.  For a fixed integer $m>1$, let $\omega_m$ denotes the $m$-th root of unity $\exp\left(\frac{2\pi \text{i}}{m}\right)$.  Then, the polynomial $$\sum_{\left(a_0,a_1,\ldots,a_r\right)\in\{0,1,\ldots,m-1\}^{r+1}}\,\omega_m^{q\,\sum_{i=0}^r\,a_i}\,\left(\sum_{i=0}^r\,a_ix_i\right)^l\tag{*}$$ vanishes identically in $\mathbb{C}\left[x_0,x_1,\ldots,x_r\right]$ for every $l=0,1,2,\ldots,r$ and $q=1,2,\ldots,m-1$.  (Here, $0^0$ is interpreted as $1$.)

Using the Multinomial Theorem,
$$\left(\sum_{i=0}^r\,a_ix_i\right)^l=\sum_{\substack{{k_0,k_1,\ldots,k_r\in\mathbb{Z}_{\geq0}}\\{k_0+k_1+\ldots+k_r=l}}}\,\binom{l}{k_0,k_1,\ldots,k_r}\,\prod_{i=0}^r\,a_i^{k_i}x_i^{k_i}\,.$$
Since $l\leq r$, $k_j=0$ for some $j$ for any monomial $\prod_{i=0}^r\,x_i^{k_i}$ in the sum above.  Hence, 
$$\sum_{a_j\in\{0,1,2,\ldots,m-1\}}\,\omega_m^{q\,a_j}\,\prod_{i=0}^r\,a_i^{k_i}x_i^{k_i}=0\,.$$
Consequently, the polynomial (*) is the zero polynomial.

Corollary:  Let $m>1$ and $r\geq 0$ be integers.  Then, the polynomial $$\sum_{\left(a_0,a_1,\ldots,a_r\right)\in\{0,1,\ldots,m-1\}^{r+1}}\,\omega_m^{q\,\sum_{i=0}^r\,a_i}\,\left(x+\sum_{i=0}^r\,a_ix_i\right)^l$$ vanishes identically in $\mathbb{C}\left[x,x_0,x_1,\ldots,x_r\right]$ for each $l=0,1,2,\ldots,r$ and $q=1,2,\ldots,m-1$.

The corollary is an immediate consequence of the lemma.  A proof consists of the binomial expansion $$\left(x+\sum_{i=0}^r\,a_ix_i\right)^l=\sum_{t=0}^l\,\binom{l}{t}\,x^{l-t}\,\left(\sum_{i=0}^r\,a_ix_i\right)^t\,.$$

Hint:  Now, apply the corollary above with $m=2$, $q=1$, $x=n$, $x_i=2^i$ for $i=0,1,2,\ldots,r$, $r=5$, and $l=5$.


This alternative solution is inspired by Kelenner's method.  We work in the polynomial ring $\mathbb{Q}[X]$.  Note that, for $r=0,1,2,\ldots$, we have $$P_r(X):=\sum_{i=0}^{2^{r+1}-1}\,f_i\,X^i=\prod_{j=0}^{r}\,\left(1-X^{2^r}\right)\,.$$
Denote by $D$ the operator $X\,\frac{\text{d}}{\text{d}X}$.  Since $P_r(X)$ is divisible by $(X-1)^{r+1}$, it follows that $D^lP_r(1)=0$ for $l=0,1,2,\ldots,r$.  However, this means
$$\sum_{i=0}^{2^{r+1}-1}\,f_i\,i^l=D^lP_r(1)=0$$
if $l$ is a nonnegative integer less than or equal to $r$.  Thus, for each $l=0,1,2,\ldots,r$,
$$\sum_{i=0}^{2^{r+1}-1}\,f_i\,(X+i)^l=\sum_{i=0}^{2^{r+1}-1}\,f_i\,\sum_{t=0}^l\,\binom{l}{t}\,X^{l-t}\,i^t=\sum_{t=0}^l\,\binom{l}{t}\,X^{l-t}\,\sum_{i=0}^{2^{r+1}-1}\,f_i\,i^t=0\,.$$
A: A tentative for another solution.
Note first that you have not to take the sum of the digits modulo $2$, as we consider $(-1)^{s(i)}$ where $s(i)$ is the sum of the digits in base $2$ of $i$. Note also that we need only to show that $\sum_{i=0}^{63} (-1)^{s(i)}i^k$ is zero for $k=0,1\cdots 5$. Now let $i=a_0+a_12+a_2 2^2+a_3 2^3+a_4 2^4+a_5 2^5$ with $a_k\in \{0,1\}$. If we note $b_k=1-a_k\in \{0,1\}$, we get easily that $b_0+b_1 2+b_2 2^2+b_3 2^3+b_4 2^4+b_5 2^5=63-i$. Hence $s(63-i)=6-s(i)$,and $(-1)^{s(63-i)}=(-1)^{s(i)}$. Now Put $S_k=\sum_{i=0}^{63}(-1)^{s(i)}i^k$. By the above, we have $S_k=\sum_{i=0}^{63}(-1)^{s(i)}(63-i)^k$, hence this is the coefficient of $x^{63}$ in the product $F_k(x)=(\sum_{l=0}^{63}(-1)^{s(l)}x^l)(\sum_{j\geq 0} j^kx^j)$. Now using the binary expansion, we get $\sum_{i=0}^{63}(-1)^{s(l)}x^l=(1-x)(1-x^2)(1-x^4)\cdots (1-x^{32})$. It is easy to see by induction that $\sum_{j\geq 0} j^kx^j=\frac{P_k(x)}{(1-x)^{k+1}}$ where $P_k$ is a polynomial of degree at most $k$ (Use that $\sum_{j\geq 0} j^kx^j=(x\frac{d}{dx})^k(\frac{1}{1-x}$)) Now we see, as $(1-x)^6$ is in factor in $(1-x)(1-x^2)(1-x^4)\cdots (1-x^{32})$, that $F_k(x)$ is a polynomial, of degree at most $ 2^6-1+{\rm degree}(P_k)-k-1=2^6-2=62$. Hence the coefficient of $x^{63}$ is zero.    
