I came across this question in my weekly test paper and didn't know how to tackle it. The question goes as follows: $a_0,a_1,a_2....a_{34}$ are the coefficients of $x^0,x^1,x^2......x^{34}$ of the polynomial obtained on opening the parenthesis of $(1+x+x^2)^{17}$, then which is true?
$1)~a_1+a_2+.......+ a_{34}$ is even;
$2)~a_1=17$;
$3)~a_{33}=17$;
$4)~a_2=153$.
(One or more is correct)
Here is some info regarding (2) - (4)
Ad (2): $a_1$ is the coefficient of $x^1$ in $(1+x+x^2)^{17}$. Since there are $17$ factors of the form $1+x+x^2$ we have to choose precisely from one factor $x$ whereas from all other factors we have to choose $1$. This can be done in \begin{align*} a_1=\binom{17}{1}=17 \end{align*} different ways.
Ad (3): The polynomial $1+x+x^2=x\left(\frac{1}{x}+1+x\right)$ has a nice symmetrical structure. This implies the coefficient of $x^k$ in $(1+x+x^2)^{17}$ is also the coefficient of $x^{34-k}$ with $0\leq k \leq 34$. We conclude \begin{align*} a_{33}=a_{34-1}=a_1=17 \end{align*}
Ad (4): $a_2$ is the coefficient of $x^2$ in $(1+x+x^2)^{17}$. In order to get $x^2$ from $17$ factors of the form $1+x+x^2$ we have to choose one factor $x^2$ which can be done in $\binom{17}{1}$ different ways and $1$ from the remaining $16$ factors, or we have to choose two factors $x$ which can be done in $\binom{17}{2}$ different ways and $1$ from the remaining $15$ factors. We conclude \begin{align*} a_2=\binom{17}{1}+\binom{17}{2}=17+\frac{1}{2}\cdot 17\cdot 16=153 \end{align*}
I have a partial response: the sum listed in (1) is almost the same as plugging in $x=1$ to the polynomial. The only difference is that you will not have the constant term $a_0=1$. Let $f(x)$ be your polynomial. Then $f(1)=3^{17}$ which is an odd number. Your sum is \begin{align*} \sum_{j=1}^{34}a_j&=f(1)-a_0\\ &=3^{17}-1 \end{align*} The sum is even!
Hint: if you plug $x=1$ in, you will get the sum of the coefficients.
The coefficients may be represented using Viete's formulas.