Step 1: Understanding the Concept:
The problem asks for the sum of a series where each term is itself the sum of consecutive odd numbers. We need to find the general term \( T_n \) and then sum it from \( n=1 \) to \( n=10 \).
Step 2: Key Formula or Approach:
1. The sum of the first \( n \) odd numbers is \( n^2 \).
2. The sum of the first \( k \) squares is given by \( \sum_{n=1}^{k} n^2 = \frac{k(k+1)(2k+1)}{6} \).
Step 3: Detailed Explanation:
Let's analyze the terms of the series:
1st term (\( T_1 \)): \( 1 = 1^2 \)
2nd term (\( T_2 \)): \( 1+3 = 4 = 2^2 \)
3rd term (\( T_3 \)): \( 1+3+5 = 9 = 3^2 \)
4th term (\( T_4 \)): \( 1+3+5+7 = 16 = 4^2 \)
From this pattern, the \( n \)-th term is the sum of the first \( n \) odd natural numbers, which is equal to \( n^2 \).
\[ T_n = n^2 \]
We need to find the sum of the first 10 terms:
\[ S_{10} = \sum_{n=1}^{10} T_n = \sum_{n=1}^{10} n^2 \]
Using the formula for the sum of squares:
\[ \sum_{n=1}^{k} n^2 = \frac{k(k+1)(2k+1)}{6} \]
Substitute \( k = 10 \):
\[ S_{10} = \frac{10(10+1)(2(10)+1)}{6} \]
\[ S_{10} = \frac{10 \cdot 11 \cdot 21}{6} \]
\[ S_{10} = \frac{10}{2} \cdot 11 \cdot \frac{21}{3} \]
\[ S_{10} = 5 \cdot 11 \cdot 7 \]
\[ S_{10} = 55 \cdot 7 \]
\[ S_{10} = 385 \]
Step 4: Final Answer:
The sum is 385.