Polynomial Long Division

Given that $P(x) = \dfrac {x^{26}-26x+25}{x^{24}+2x^{23}+3x^{22}+4x^{21}+\cdots + 24x+25}$ , this means that $P(x)$ has a degree of 2. Let $P(x)=ax^2 + bx + c$ . Now consider:

$\begin{aligned} P(0) & = \frac {25}{25} = 1 & \small \color{#3D99F6} \implies P(0) = c \implies c = 1 \\ P(1) & = \frac {1-26+25}{1+2+3+\cdots + 25} = 0 & \small \color{#3D99F6} \implies P(1) = a + b + 1 = 0 \quad ...(1) \\ P(-1) & = \frac {1+26+25}{\underbrace{1-2}_{=-1}+\underbrace{3-4}_{=-1}+\cdots + \underbrace{23-24}_{=-1} + 25} \\ & = \frac {52}{-12+25} = \frac {52}{13} = 4 & \small \color{#3D99F6} \implies P(-1) = a - b + 1 = 4 \quad ...(2) \end{aligned}$

From $(1)+(2): \ 2a+2=4 \implies a = 1$ and $(1): \ 1+b+1 = 0 \implies b = -2$ . Therefore, $P(x) = x^2 - 2x + 1$ and the sum of coefficients is $1-2+1=\boxed 0$ .

---

Let us prove that $P(x) = \dfrac {x^{26}-26x+25}S = x^2-2x+1$ , where $S= x^{24}+2x^{23}+3x^{22}+4x^{21}+\cdots + 24x+25$ , by showing $(x^2 - 2x+1)S = x^{26} - 26x + 25$ .

$\begin{aligned} S & = x^{24}+2x^{23}+3x^{22}+4x^{21}+\cdots + 24x+25 & \small \color{#3D99F6} ...(1) \\ xS & = x^{25}+2x^{24}+3x^{23}+4x^{22}+\cdots + 24x^2+25x & \small \color{#3D99F6} ...(2) \\ x^2S & = x^{26}+2x^{25}+3x^{24}+4x^{23}+\cdots + 24x^3+25x^2 & \small \color{#3D99F6} ...(3) \\ (x^2 - x)S & = x^{26}+x^{25}+x^{24}+x^{23}+\cdots + x^2 - 25x & \small \color{#3D99F6} ...(3)-(2) = (4) \\ (-x+1)S & = -x^{25}-x^{24}-x^{23}-x^{22}-\cdots - x + 25 & \small \color{#3D99F6} ...(1)-(2) = (5) \\ (x^2 - 2x+1)S & = x^{26}-26x+25 & \small \color{#3D99F6} ...(4)+(5) \end{aligned}$

Proving $P(x) = x^2-2x+1$ .

The sum of coefficients of any polynomial P(x) will be equal to P(1).As it would add only the coefficients and will not be affected by the index of variable. Putting x=1 in the whole division, in numerator,we get; 1^2-26+25=0; Thus,P(1)=0 and the sum of coefficients is zero.

Working from the geometric series formula gives us that: $\frac{x^{26}-1}{x-1} = x^{25} + x^{24} + x^{23} + \cdots + x + 1$

$x^{26}-26x+25 = x^{26}-1-26x+26 = x^{26}-1 -26(x-1)$

$\frac{x^{26}-26x+25}{x-1} = x^{25} + x^{24} + x^{23} + \cdots + x + 1 -26$

$\frac{x^{26}-26x+25}{x-1} = x^{25} + x^{24} + x^{23} + \cdots + x - 25$

This gives us the idea to try multiplying the denominator by $x-1$ .

$(x-1)(x^{24} + 2x^{23} + 3x^{22} + 4x^{21} + \cdots + 24x + 25) = (x^{25} + 2x^{24} + 3x^{23} + 4x^{22} + \cdots + 24x^2 + 25x) - (x^{24} + 2x^{23} + 3x^{22} + 4x^{21} + \cdots + 24x + 25)$

$= x^{25} + (2x^{24}-x^{24}) + (3x^{23}-2x^{23}) + (4x^{22}-3x^{22}) + \cdots + (25x-24x) - 25$

$= x^{25} + x^{24} + x^{23} + \cdots + x - 25$

Now we combine our results: $\frac{x^{26}-26x+25}{x-1} = x^{25} + x^{24} + x^{23} + \cdots + x - 25$

$(x-1)(x^{24} + 2x^{23} + 3x^{22} + 4x^{21} + \cdots + 24x + 25) = x^{25} + x^{24} + x^{23} + \cdots + x - 25$

$\frac{x^{26}-26x+25}{x-1} = (x-1)(x^{24} + 2x^{23} + 3x^{22} + 4x^{21} + \cdots + 24x + 25)$

$\frac{x^{26}-26x+25}{x^{24} + 2x^{23} + 3x^{22} + 4x^{21} + \cdots + 24x + 25} = (x-1)^2 = x^2 - 2x + 1$ So the sum of the coefficients is $\boxed{0}$ .

Just doing the polynomial division: $\frac{x^{26}-26 x+25}{\sum _{i=1}^{25} i\ x^{25-i}}=x^2-2 x+1\ \text{remainder}\ 0$ .

We can see that the question actually asks for the value of Q(1)....... Here Q(x) is the quotient obtained by the expression given...........Therefore, simply putting x=1 in the expression gives the desired answer!!!!!