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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. $80751$
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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}} $

Reference: https://brilliant.org/wiki/sum-of-n-n2-or-n3/

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