Find the sum up to 25 terms of the following sequence.

1+9+29+ 67 +

The sequence provided appears to be a quadratic sequence. To find the sum of the first 25 terms, we first need to determine the general formula for the \(n\)-th term of the sequence, which takes the form \( T_n = an^2 + bn + c \).

### Step 1: Find the general term

Given the first few terms:

\[
T_1 = 1, \quad T_2 = 9, \quad T_3 = 29, \quad T_4 = 67
\]

We will first compute the first differences:

\[
9 - 1 = 8, \quad 29 - 9 = 20, \quad 67 - 29 = 38
\]

Then, compute the second differences:

\[
20 - 8 = 12, \quad 38 - 20 = 18
\]

Since the second differences are constant, the sequence is quadratic.

### Step 2: Set up the system of equations

We assume the \(n\)-th term is given by \( T_n = an^2 + bn + c \).

Using the known values of the terms:

\[
\begin{aligned}
T_1 &= a(1)^2 + b(1) + c = 1, \\
T_2 &= a(2)^2 + b(2) + c = 9, \\
T_3 &= a(3)^2 + b(3) + c = 29.
\end{aligned}
\]

This gives the system of equations:

\[
\begin{aligned}
a + b + c &= 1, \\
4a + 2b + c &= 9, \\
9a + 3b + c &= 29.
\end{aligned}
\]

### Step 3: Solve the system of equations

Subtract the first equation from the second and the second from the third:

\[
\begin{aligned}
(4a + 2b + c) - (a + b + c) &= 9 - 1 \implies 3a + b = 8, \\
(9a + 3b + c) - (4a + 2b + c) &= 29 - 9 \implies 5a + b = 20.
\end{aligned}
\]

Now subtract these two equations:

\[
(5a + b) - (3a + b) = 20 - 8 \implies 2a = 12 \implies a = 6.
\]

Substitute \(a = 6\) into \(3a + b = 8\):

\[
3(6) + b = 8 \implies 18 + b = 8 \implies b = -10.
\]

Substitute \(a = 6\) and \(b = -10\) into \(a + b + c = 1\):

\[
6 - 10 + c = 1 \implies c = 5.
\]

Thus, the general term is:

\[
T_n = 6n^2 - 10n + 5.
\]

### Step 4: Find the sum of the first 25 terms

The sum of the first \(n\) terms is:

\[
S_n = \sum_{n=1}^{25} (6n^2 - 10n + 5).
\]

We can break this into three separate sums:

\[
S_{25} = 6 \sum_{n=1}^{25} n^2 - 10 \sum_{n=1}^{25} n + 5 \times 25.
\]

Use the known formulas for the sums of \(n^2\) and \(n\):

\[
\sum_{n=1}^{25} n^2 = \frac{25(25+1)(2(25)+1)}{6} = 5525, \quad \sum_{n=1}^{25} n = \frac{25(25+1)}{2} = 325.
\]

Now substitute these into the expression for \(S_{25}\):

\[
S_{25} = 6(5525) - 10(325) + 5 \times 25 = 33150 - 3250 + 125 = 30025.
\]

### Final Answer:
The sum of the first 25 terms is \( \boxed{30025} \).

Would you like further details or have any questions?

Here are five related questions for deeper exploration:
1. What is the sum of the first 50 terms of the same sequence?
2. How would the approach change if the sequence were arithmetic instead of quadratic?
3. How can you verify that the second differences remain constant?
4. Can you derive the formula for the sum of the first \(n\) terms of a quadratic sequence?
5. How do sums of cubic sequences compare to sums of quadratic ones?

**Tip:** Identifying the type of sequence early on (arithmetic, geometric, quadratic, etc.) can save a lot of time when solving problems involving sums.

Find the sum up to 25 terms of the following sequence: 1+9+29+67+...

Learn how to find the sum of the first 25 terms of a quadratic sequence. The solution involves identifying the general term of the sequence, using quadratic formulas, and applying summation techniques for n^2 and n terms. Suitable for students in Grades 10-12.

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