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

Look at the terms: $7 \times 11 + 11 \times 15 + 15 \times 19 + \dots + 95 \times 99$

Notice each term shares a number with the next term:

• $7 \times 11$ then $11 \times 15$

• $11 \times 15$ then $15 \times 19$

The numbers go: $7, 11, 15, 19, 23, 27\ldots$ (adding $4$ each time)

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Each number in our sequence = $4n + 3$ where $n = 1, 2, 3\ldots$

When $n = 1$: $4(1) + 3 = 7$

When $n = 2$: $4(2) + 3 = 11$

So each term looks like: $(4n + 3) \times (4(n+1) + 3) = (4n + 3)(4n + 7)$

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Last number is $95$, so: $4n + 3 = 95$

$4n = 92$

$n = 23$

We have $23$ terms total.

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$\begin{aligned} (4n + 3)(4n + 7) &= 4n(4n + 7) + 3(4n + 7) \\ &= 16n^2 + 28n + 12n + 21 \\ &= 16n^2 + 40n + 21 \end{aligned}$

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Our sum = $\sum_{n=1}^{23} (16n^2 + 40n + 21)$

Using the formulas:

$\sum_{n=1}^{N} n^2 = \dfrac{N(N+1)(2N+1)}{6}$

$\sum_{n=1}^{N} n = \dfrac{N(N+1)}{2}$

$\sum_{n=1}^{N} 1 = N$

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For $N = 23$:

$\sum_{n=1}^{23} n^2 = \dfrac{23 \times 24 \times 47}{6} = \dfrac{25944}{6} = 4324$

$\sum_{n=1}^{23} n = \dfrac{23 \times 24}{2} = 276$

$\sum_{n=1}^{23} 1 = 23$

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$\begin{aligned} \text{Sum} &= 16(4324) + 40(276) + 21(23) \\ &= 69184 + 11040 + 483 \\ &= 80707 \end{aligned}$