Question

The value of the sum 7 x 11 + 11 x 15 + 15 x 19 + ...+ 95 x 99 is

• A. 80707
• B. 80773
• C. 80730
• D. 80751

Answer 1

The Correct Option is A

The correct answer is (A): 

The \( n^\text{th} \) term of the series is: \[ t_n = (4n - 3)(4n - 7) \]

Expanding this: \[ t_n = 16n^2 - 28n - 21 \]

The sum of the first \( n \) terms is: \[ \sum t_n = \sum (16n^2 - 28n - 21) = 16 \sum n^2 - 28 \sum n - 21n \]

Using formulae: \[ \sum n^2 = \frac{n(n+1)(2n+1)}{6}, \quad \sum n = \frac{n(n+1)}{2} \]

Therefore: \[ \sum t_n = 16 \cdot \frac{n(n+1)(2n+1)}{6} - 28 \cdot \frac{n(n+1)}{2} - 21n \]

The series terms are \( 7, 11, 15, \ldots, 95 \), an arithmetic progression with common difference 4.

Number of terms \( n \) is: \[ n = \frac{95 - 7}{4} + 1 = \frac{88}{4} + 1 = 22 + 1 = 23 \]

Now plug in \( n = 23 \):

\[ \sum t_n = 16 \cdot \frac{23 \cdot 24 \cdot 47}{6} - 28 \cdot \frac{23 \cdot 24}{2} - 21 \cdot 23 \]

Compute step-by-step:

• \( \frac{23 \cdot 24 \cdot 47}{6} = 4332 \)
• \( 16 \cdot 4332 = 69312 \)
• \( \frac{23 \cdot 24}{2} = 276 \)
• \( 28 \cdot 276 = 7728 \)
• \( 21 \cdot 23 = 483 \)

Final sum: \[ \sum t_n = 69312 - 7728 - 483 = \boxed{80707} \]