Coefficient of $x^{50}$ in the polynomial. What is the coefficient of $x^{50}$ in the polynomial 
$[x+\binom{50}{0}][x+3\binom{50}{1}][x+5\binom{50}{2}].....[x+(2n+1)\binom{50}{50}]$
The solution given in one of the books was ,
We have ,
$[x+\binom{50}{0}][x+3\binom{50}{1}][x+5\binom{50}{2}].....[x+(2n+1)\binom{50}{50}]$
$=x^{51} +x^{50}[\binom{50}{0}+ 3\binom{50}{1} + 5\binom{50}{2} .... 101\binom{50}{50}$]
And from here they have found out the coefficient of $x^{50}$ 
However I am not able to understand, how is the coefficient of $x^{50}$ the sum of all of the binomial coefficients in the polynomial? Is there a general method to determine the coefficients of the second or the third highest degree variable of any polynomial using binomial theorem?
 A: You obtain the power $x^{50}$ when in your $51$ terms, you pick $x$ fifty times.
Firstly, you pick the constant from the last term, then from the second to last term etc until you pick the constant from the first term. You have to add all those together so you obtain
$$x^{50} + x^{50}3\binom{50}{1} +\ldots + x^{50}(2n+1) $$
A: To get one $x^{50}$ term, you take the $x$ term (coefficient: one) from fifty sets of parentheses and a single constant term from the single remaining set of parentheses. Adding all of these up gives you the total coefficient of $x^{50}$.
As a simple analogy, consider the $x^2$ term in $(x+1)(x+3)(x+5)$.
You take the $x$ term from $(x+1)$, multiply it by the $x$ term from $(x+3)$ and finally by the constant term from $(x+5)$. That gives you $(x)(x)(5) = 5x^2$. Rotating through the possible combinations here, you get $5x^2 + 3x^2 + x^2 = 9x^2$. The same method is at play here.
A: There is in fact a formula for the coefficients. 
We observe
$$
(x+a_1)(x+a_2)=x^2+(a_1+a_2)x+a_1a_2\\
(x+a_1)(x+a_2)(x+a_3)=x^3+(a_1+a_2+a_3)x^2+(a_1a_2+a_1a_3+a_2a_3)x+a_1a_2a_3\\
\vdots
$$
We see that the $n$-th coefficient ist always $1$, the $(n-1)$-th coefficient ist always the sum of all $a_1,\ldots,a_n$, the $(n-2)$-th coeffient is the product of two different factors from $a_1,\ldots,a_n$ and so on, such that the last coeffient is just the product of all $a_1,\ldots,a_n$.
We deduce:
The $k$-th coefficient of 
$$
(x+a_1)(x+a_1)\cdot\ldots\cdot(x+a_n)
$$
is given by
$$
\sum_{\substack{m_1,\ldots,m_{n-k}=1\\m_i\neq m_j\text{ for }i\neq j}}^na_{m_1}\cdot\ldots\cdot a_{m_{n-k}}.
$$
(Proof by induction).
Normally you are interested in the $n$-th, $(n-1)$-th  and $0$-th coefficient, which is given by
$$
n\text{-th coefficient: }1\\
(n-1)\text{-th coefficient: }\sum_{m=1}^na_m\\
0\text{-th coefficient: }\prod_{m=1}^na_m\\
$$
A: Let's take a smaller example and then generalise.
$[x+\binom{3}{0}][x+3\binom{3}{1}][x+5\binom{3}{2}][x+(7)\binom{3}{3}]$
$\quad =x^4$
$\quad + x^3[\binom30+3\binom31+5\binom32+7\binom33]$
$\quad+x^2[\binom303\binom31+\binom305\binom32+\binom307\binom33+3\binom315\binom32+3\binom317\binom33+5\binom327\binom33]$
$\quad +x[\binom303\binom315\binom32+\binom303\binom317\binom33+\binom305\binom327\binom33+3\binom315\binom327\binom33]$
$\quad + \binom303\binom315\binom327\binom33$
So, for the first term, we just multiply each $x$ with the other $x$. In the second term, we multiply all but one $x$'s and so instead of multiplying with another $x$ in the remaining bracket, we multiply with the constant part in the bracket, that's why we get a sum of each binomial coefficient. Next, for the next power of $x$, we multiply all but two x's and then multiply with the remaining two constant parts in each bracket, and so we get a sum of each pairwise multiplication of each constant term. In the next term, all but three x's are taken and each triplet of the constant part of in the remaining three brackets. And so on...
