solve a,b,c,d,e are real numbers such that

a+4b+ac+16d+25e=1

4a+9b+16c+25d+36e=8

9a+16b+25c+36d+49e=23

Then a+b+c+d+e is

We are given the system of linear equations:

\[
\begin{aligned}
1. & \quad a + 4b + ac + 16d + 25e = 1 \\
2. & \quad 4a + 9b + 16c + 25d + 36e = 8 \\
3. & \quad 9a + 16b + 25c + 36d + 49e = 23 \\
\end{aligned}
\]

These equations can be rewritten in matrix form as:

\[
\begin{pmatrix}
1 & 4 & 9 & 16 & 25 \\
4 & 9 & 16 & 25 & 36 \\
9 & 16 & 25 & 36 & 49 \\
\end{pmatrix}
\begin{pmatrix}
a \\
b \\
c \\
d \\
e \\
\end{pmatrix}
=
\begin{pmatrix}
1 \\
8 \\
23 \\
\end{pmatrix}
\]

Notice that the coefficients of the equations correspond to squares of consecutive integers (1, 2, 3, 4, 5, etc.). This is a common pattern in problems involving sequences of quadratic forms.

Let’s examine the structure of the matrix and solve for \( a + b + c + d + e \).

### Step 1: Test if the sum is consistent with a polynomial sequence.

Assume the variables \( a, b, c, d, e \) correspond to a polynomial sequence of degree 2:

\[
S(n) = A n^2 + B n + C
\]

Given:

\[
\begin{aligned}
S(1) & = A(1)^2 + B(1) + C = A + B + C = 1 \\
S(2) & = A(4) + B(2) + C = 4A + 2B + C = 8 \\
S(3) & = A(9) + B(3) + C = 9A + 3B + C = 23 \\
\end{aligned}
\]

### Step 2: Solve the polynomial coefficients.

We have the system:

\[
\begin{aligned}
A + B + C & = 1 \\
4A + 2B + C & = 8 \\
9A + 3B + C & = 23 \\
\end{aligned}
\]

Subtract the first equation from the second:

\[
(4A + 2B + C) - (A + B + C) = 8 - 1 \\
3A + B = 7 \quad \text{(Equation 4)}
\]

Subtract the second equation from the third:

\[
(9A + 3B + C) - (4A + 2B + C) = 23 - 8 \\
5A + B = 15 \quad \text{(Equation 5)}
\]

Subtract Equation 4 from Equation 5:

\[
(5A + B) - (3A + B) = 15 - 7 \\
2A = 8 \quad \Rightarrow \quad A = 4
\]

Substitute \( A = 4 \) into Equation 4:

\[
3(4) + B = 7 \quad \Rightarrow \quad B = -5
\]

Substitute \( A = 4 \) and \( B = -5 \) into the first original equation:

\[
4 - 5 + C = 1 \quad \Rightarrow \quad C = 2
\]

Thus, \( A = 4 \), \( B = -5 \), and \( C = 2 \).

### Step 3: Calculate the sum \( a + b + c + d + e \):

Given the pattern:

\[
S(n) = 4n^2 - 5n + 2
\]

For \( n = 1, 2, 3, 4, 5 \):

\[
S(1) = 4(1) - 5(1) + 2 = 1, \quad S(2) = 8, \quad S(3) = 23, \quad S(4) = 50, \quad S(5) = 89
\]

Hence:

\[
a + b + c + d + e = S(1) + S(2) + S(3) + S(4) + S(5) = 1 + 8 + 23 + 50 + 89 = \boxed{5}
\]

Learn how to solve a system of linear equations involving polynomial sequences. This problem explores matrix operations and polynomial coefficients, providing a step-by-step solution. Suitable for advanced high school students (Grades 10-12).

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