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The value of the sum $7 \times 11 + 11 \times 15 + 15 \times 19 + \dots$ + $95 \times 99$ is

1. $80707$
2. $80773$
3. $80730$
4. $80751$

Let $S = 7 \times 11 + 11 \times 15 + 15 \times 19 + \dots + 95 \times 99$

We can write first series as :

$S_{1} = \underbrace{7 + 11 + 15 + \dots + 95}_{\text{Arithmetic Progression (AP)}}$

Here, first term $a = 7,$  common difference $d = 11 – 7 = 4,$  last term $l = 95$

The $n^{th}$ term of the series $t_{n} = l = a+(n-1)d \;,$ where $n =$ number of terms

$\Rightarrow 95 = 7+(n-1)4$

$\Rightarrow 4n – 4 + 7 = 95$

$\Rightarrow 4n + 3 = 95$

$\Rightarrow 4n = 92$

$\Rightarrow \boxed{ n= 23}$

The $n^{th}$ term of the series $\boxed{t_{n} = 4n + 3}$

Similarly, we can write second series as :

$S_{2} = \underbrace{11 + 15 + 19 + \dots + 99}_{\text{Arithmetic Progression (AP)}}$

Here, $a = 11, \; d = 15 – 11 = 4, \; l = 99, \; n = 23$

The $n^{th}$ term of the series $t_{n} = l = a + (n-1)d$

$\Rightarrow t_{n} = 11 + (n-1)4$

$\Rightarrow t_{n} = 11 + 4n – 4$

$\Rightarrow \boxed{t_{n} = 4n + 7}$

Now, we can write  $T_{n} = (4n+3)(4n+7) \;, \text{where} \; n = 1, 2, \dots, 23$

$\Rightarrow T_{n} = 16n^{2} + 28n + 12n + 21$

$\Rightarrow T_{n} = 16n^{2} + 40n + 21$

$\Rightarrow \sum T_{n} = \sum (16n^{2} + 40n + 21)$

$\Rightarrow S_{n} = 16 \sum n^{2} + 40 \sum n + 21 \sum 1$

$\Rightarrow S_{n} = 16 \left[ \frac{n(n+1)(2n+1)}{6} \right] + 40 \left[ \frac{n(n+1)}{2} \right] + 21n$

The sum of  the series $S = 16 \left[\frac{(23)(24)(47)}{6} \right] + 40 \left[ \frac{(23)(24)}{2} \right] + 21 \times 23$

$\Rightarrow S = 69184 + 11040 + 483$

$\Rightarrow \boxed{S = 80707}$

Correct Answer $: \text{A}$

$\textbf{PS :}$

If $S_{n} = 1 + 2 + 3 + \dots + n$

$\Rightarrow S_{n} = \displaystyle{}\sum_{k=1}^{n} k$

$\Rightarrow \boxed{S_{n} = \frac{n(n+1)}{2}}$

If $S_{n} = 1^{2} + 2^{2} + 3^{2} + \dots + n^{2}$

$\Rightarrow S_{n} = \displaystyle{}\sum_{k=1}^{n}k^{2}$

$\Rightarrow \boxed{S_{n} = \frac{n(n+1)(2n+1)}{6}}$

If $S_{n} = 1^{3} + 2^{3} + 3^{3} + \dots + n^{3}$

$\Rightarrow S_{n} = \displaystyle{}\sum_{k=1}^{n} k^{3}$

$\Rightarrow \boxed{S_{n} = \left[ \frac{n(n+1)}{2} \right]^{2}}$

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