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Explanation:

1 × 3 + 5 × 7 + 9 × 11 + 

T₁ = 1 × 3
T₂ = 5 × 7 and so on

In each of these terms, the first number forms an AP whose first term is 1 and common difference is 4.
∴ first number of n^th term = 1 + (n - 1) × 4 = 4n - 3

In each of these terms, the second number forms an AP whose first term is 3 and common difference is 4.
∴ second number of n^th term = 3 + (n - 1) × 4 = 4n - 1

⇒ T_n = (4n - 3) × (4n - 1)
⇒ T_n = 16n² - 16n + 3

∴ $\underset{n=1}{\overset{10}{\sum T_n}}$ = 16(1² + 2² + 3² + ... + 10²) + 16(1 + 2 + 3 + 3 + ... + 10) + (3 + 3 + 3 + ... + 3)

= $16\left(\frac{10\times 11\times 21}{6}\right)$ - $16\left(\frac{10\times 11}{2}\right)$ + 30

= 6160 - 880 + 30 = 5310

Hence, 5310.