The $49^\text{th}$ Coefficient

Obviously, the product above is equal to $(n)(n+1)(n+2) \cdots (n+52)$

Let $P(x) = (n)(n+1)(n+2) \cdots (n+52)$ Then, obviously, $P(x)$ has roots $-1, -2, -3, \cdots -52$

First, let us find the sums of all the roots, the sums of the squares of the roots and the sums of the cubes of the roots.

Obviously, the sum of the roots of the equation is equal to $-1-2-3 \cdots -52$ or $-(1+2+3+ \cdots 52)$

This is equal to $-\frac {(52)(53)}{2} = -1378$

Similarly, we get the sum of the squares of the roots $(-1)^2+(-2)^2+(-3)^2 + \cdots + (-52)^2= 1^2+2^2+3^2+\cdots +52^2 = \frac{(52)(53)(105)}{6} = 48230$

Also, the sum of the cubes of the roots are $(-1)^3+(-2)^3+(-3)^3+\cdots +(-52)^3 = -(1^3+2^3+3^3+\cdots+52^2)=-(\frac{(52)(53)}{2})^2 = -1898884$

Now, finally, we can proceed with the Newton's sums part.

Since $P(x)= x^{52} + a_{51}x^{51}+a_{50}x^{50}+a_{49}x^{49}+\cdots + a_0$ , then by Newton's sums,

$S_1+a_{51} = 0$

$S_2+S_1a_{51}+2a_{50}= 0$

$S_3+S_2a_{51}+S_1a_{50} + 3a_{49}= 0$

Where $S_n = \displaystyle \sum_{i=1}^{52} x_i^n$

Solving for the variables $a_{51}, a_{50}, a_{49}$ ,

$S_1+a_{51} = 0$

$a_{51} = 1378$

$S_2+S_1a_{51}+2a_{50}= 0$

$a_{50} = 925327$

$S_3+S_2a_{51}+S_1a_{50} + 3a_{49}= 0$

$a_{49} = 403512850$

Where the equations above were solved by simple substitution and use of calculator.

So, we get $a_{49} = \boxed{403512850}$

I found a solution I started using a while back which solves all problems of this type really easily.

Consider $\prod_{n=1}^{52} (1+yn)$ . It should be clear that the coefficient of $y^k$ is equal to the coefficient of $x^{52-k}$ . An expansion of each of the expressions shows this. Hence, we should look for the coefficient of $y^3$ in the said expression. This proof uses the MacLaurin expansions of $\ln(1+x)$ and of $e^x$ .

Now. Consider the following: $\begin{aligned} \prod_{n=1}^{52} (1+yn) &= \exp\left({\ln {\prod_{n=1}^{52} (1+yn)}}\right) \\ &= \exp\left({\ln (1+y) + \ln(1+2y) + \dots + \ln(1+52y)}\right) \\ &= \exp\left({\left(y - \frac{y^2}{2} + \frac{y^3}{3} - \dots\right) + \left(2y - \frac{4y^2}{2} + \frac{8y^3}{3} - \dots\right) + \dots + \left(52y - \frac{52^2y^2}{2} + \frac{52^3y^3}{3} - \dots\right)}\right) \\ &= \exp\left({\sum_{n=1}^{52} ny - \frac{1}{2}\sum_{n=1}^{52} n^2y^2 + \frac{1}{3}\sum_{n=1}^{52} n^3 y^3 - \dots }\right) \end{aligned}$

Define $S_k = \frac{1}{k}\sum_{n=1}^{52}n^k$ .

$\begin{aligned} \prod_{n=1}^{52} (1+yn) &= \exp\left(S_1y - S_2y^2 + S_3y^3 - \dots\right) \\ &= e^{S_1 y} \times e^{-S_2 y^2} \times e^{S_3 y^3} \times \dots \\ &= (1 + (S_1 y) + \frac{(S_1 y)^2}{2} + \frac{(S_1 y)^3}{6} + \dots) \times (1 + (-S_2 y^2) + \frac{(-S_2 y^2)^2}{2} + \frac{(-S_2 y^2)^3}{6} + \dots) \times (1 + (S_3 y^3) + \frac{(S_3 y^3)^2}{2} + \frac{(S_3 y^3)^3}{6} + \dots) \times \dots \\ &= 1 + (S_1) y + \left(\frac{S_1^2}{2} - S_2 \right) y^2 + \left( \frac{S_1^3}{3} - S_1S_2 + S_3 \right)y^3 + \dots \end{aligned}$

And hence by comparison, the coefficient of the expression we're looking for is $\frac{S_1^3}{3} - S_1S_2 + S_3$ . By using the sum of powers, we get $S_1 = 1378, S_2 = 24115, S_3 = \frac{1898884}{3}$ and thus by direct calculation, we get our coefficient of $x^{49}$ to be $\boxed{403512850}$ .

The terms with $x^{49}$ are of the form $a\cdot b\cdot c\cdot x^{49}$ , where $1 \leq a < b < c \leq 52$ .

My strategy for adding all these products $a\cdot b\cdot c$ together is as follows:

• First, allow $a, b, c$ to range over all values from 1 to 52.

• Then subtract all triples of the form $(a, a, b)$ , $(a, b, a)$ , and $(b, a, a)$ .

• In the previous step, when $a = b$ we have subtracted the same triple three times instead of once. Therefore add back in twice all triples of the form $(a, a, a)$ .

• Now we have all triples with $a \not= b\not= c\not= a$ , but each triple in all of its six permutations. Therefore divide the result by 6.

Step 1 : Sum of all products $a\cdot b\cdot c$ . This is $\sum_{a = 1}^{52} \sum_{b = 1}^{52} \sum_{c = 1}^{52} abc = \left(\sum_{a=1}^{52} a\right) \cdot \left(\sum_{b=1}^{52} b\right) \cdot \left(\sum_{c=1}^{52} c\right) = \left(\sum_{a=1}^{52} a\right)^3;$ this is equal to $\left(\frac{52\cdot 53}2\right)^3 = 2\:616\:662\:152.$

Step 2 : Sum of all products $a\cdot a\cdot b$ . This is $\sum_{a = 1}^{52} \sum_{b = 1}^{52} a^2b = \left(\sum_{a=1}^{52} a^2\right) \cdot \left(\sum_{b=1}^{52} b\right) \\ = \left(\frac{52\cdot 53\cdot 105}6\right) \cdot \left(\frac{52\cdot 53}2\right) = 66\:460\:940.$

Step 3 : Sum of all products $a\cdot a\cdot a$ . This is $\sum_{a = 1}^{52} a^3 = \left(\sum_{a = 1}^{52} a\right)^2 = \left(\frac{52\cdot 53}2\right)^2 = 1\:898\:884.$

Step 4 : Sum of all products $a\cdot b\cdot c$ with all different numbers: $2\:616\:662\:152 - 3\times 66\:460\:940 + 2\times 1\:898\:884 = 2\:421\:077\:100;$ divide by six to have sum of all products $a\cdot b\cdot c$ with $a < b < c$ : $\frac{2\:421\:077\:100}6 = \boxed{403\:512\:850}.$

Note: This can be generalized, of course. Replacing 52 by $n$ , define $S_1 = \sum_{a=1}^n a = \frac{n(n+1)}2;\\ S_2 = \sum_{a=1}^n a^2 = \frac{n(n+1)(2n+1)}6 = \frac{2n+1}3 S_1; \\ S_3 = \sum_{a=1}^n a^3 = S_1^2;$ the values for steps 1, 2, and 3 become $N_1 = S_1^3;\ \ N_2 = S_2 S_1 = \frac{2n+1}3 S_1^3;\ \ N_3 = S_1^2$ so that the answer is $N = \frac{N_1 - 3N_2 + 2N_3}6 = \tfrac16 S_1^2\cdot (S_1 - (2n+1) + 2) = \tfrac16 S_1^2 (S_1 - 2n + 1).$ In our case, $S_1 = 1378;\ \ N = \tfrac16\cdot 1378^2\cdot 1275 = 403\:512\:850.$

Start with $\begin{array}{rcl} \displaystyle F_2(N) \; = \; \sum_{1 \le r < s \le N+1} rs & = & \displaystyle \sum_{s=2}^{N+1} s \left(\sum_{r=1}^{s-1}r\right) \; = \; \frac12\sum_{s=2}^{N+1} s(s-1)s \\ & = & \displaystyle \frac12\sum_{s=1}^N s(s+1)^2 \; = \; \frac1{24}N(N+1)(N+2)(3N+5) \end{array}$ then $\begin{array}{rcl} \displaystyle F_3(N) \; = \; \sum_{1 \le r < s < t \le N+2}rst &= & \displaystyle \sum_{1 \le r < s \le N+1}rs\left(\sum_{t=s+1}^{N+2}t\right) \\ & + & \displaystyle \frac12(N+2)(N+3)F_2(N) - \frac12\sum_{1 \le r<s \le N+1} rs^2(s+1) \end{array}$ and $\begin{array}{rcl} \displaystyle \sum_{1 \le r<s \le N+1} rs^2(s+1) & = & \displaystyle\sum_{s=2}^{N+1}s^2(s+1)\left(\sum_{r=1}^{s-1}r\right) \; = \; \frac12\sum_{s=2}^{N+1}(s-1)s^3(s+1) \\ & = & \displaystyle \frac12\sum_{s=1}^N s(s+1)^3(s+2) \\ & = & \displaystyle \frac12\sum_{s=1}^N\left[ \begin{array}{l} s(s+1)(s+2)(s+3)(s+4) - 5s(s+1)(s+2)(s+3) \\+ 4s(s+1)(s+2)\end{array}\right] \\ & = & \displaystyle\frac12\left[ \begin{array}{l} \frac16N(N+1)(N+2)(N+3)(N+4)(N+5) \\- N(N+1)(N+2)(N+3)(N+4) \\+ N(N+1)(N+2)(N+3)\end{array}\right] \end{array}$ Putting all these together, we obtain $F_3(N) \; = \; \frac{1}{48}N(N+1)(N+2)^2(N+3)^2$ The answer to the question is $F_3(50) \,=\, \boxed{403512850}$ .

The coefficients of a polynomial $P(x)$ $P(x) := \sum_{k=0}^n a_kx^k=\prod_{k=1}^n (x-x_k),\qquad a_k,\: x_k\in\mathbb{C},\quad a_n=1$

satisfy a recurrence relation we can get if we rewrite the initial condition of Newton's Identities in matrix form: $\begin{pmatrix} n & S_1 & \dots & S_n\\ & \ddots & \ddots & \vdots\\ & & 1 & S_1\\ & & & 1 \end{pmatrix} \cdot \begin{pmatrix} a_0\\\vdots\\a_{n-1}\\a_n \end{pmatrix}=\begin{pmatrix} 0\\\vdots\\0\\1 \end{pmatrix},\qquad S_i=\prod_{k=1}^nx_k^i$

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Now to the task. For simplicity, let $n=52$ , so we need to find $a_{n-3}$ and only need the lower right $4\times 4$ corner of the recurrence relation containing $S_1,\:S_2,\:S_3$ . Our polynomial has the roots $x_k=-k$ and we get $\begin{aligned} S_1 &=\sum_{k=1}^n(-k)=-\frac{n(n+1)}{2}, & S_2 &=\sum_{k=1}^n (-k)^2 = \frac{n(n+1)(2n+1)}{6}, & S_3 &= \sum_{k=1}^n(-k)^3=-\frac{n^2(n+1)^2}{4} \end{aligned}$ We solve these four equations from bottom to top and obtain $\begin{aligned} a_{n-1}&=\frac{n(n+1)}{2}, & a_{n-2}&=\frac{(n-1)n(n+1)(3n+2)}{24}, & a_{n-3} &= \frac{(n-2)(n-1)n^2(n+1)^2}{48} \end{aligned}$ The special case $n=52$ leads to $a_{49}=\boxed{403512850}$