<s>[INST] 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?
 [/INST] 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.
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