If $(2n + 1) + (2n + 3) + (2n + 5) + .... + (2n + 47) = 5280$, then what is the value of $1 + 2 + 3 + ... + n$?

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The series is: $(2n + 1) + (2n + 3) + (2n + 5) + \ldots + (2n + 47)$

To find how many terms we have, we'll look at the pattern:

First term: $2n + 1$ (where the added number is 1)

Second term: $2n + 3$ (where the added number is 3)

Third term: $2n + 5$ (where the added number is 5)

Last term: $2n + 47$ (where the added number is 47)

The added numbers form the sequence: $1, 3, 5, 7, \ldots, 47$

This is the sequence of first few odd numbers. To count them:

The $k$-th odd number is $2k - 1$

So if $2k - 1 = 47$, then $k = 24$

Therefore, we have 24 terms in our series.

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Each term has $2n$ in it, so we can factor this out:

$(2n + 1) + (2n + 3) + (2n + 5) + \ldots + (2n + 47)$

$= 2n + 2n + 2n + \ldots + 2n + (1 + 3 + 5 + \ldots + 47)$

$= 24 \times 2n + (1 + 3 + 5 + \ldots + 47)$

$= 48n + (1 + 3 + 5 + \ldots + 47)$

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The sum $1 + 3 + 5 + \ldots + 47$ is the sum of the first 24 odd numbers.

There's a beautiful pattern here: the sum of the first $k$ odd numbers equals $k^2$.

Let us verify this with small examples:

First 1 odd number: $1 = 1^2$

First 2 odd numbers: $1 + 3 = 4 = 2^2$

First 3 odd numbers: $1 + 3 + 5 = 9 = 3^2$

So, $1 + 3 + 5 + \ldots + 47 = 24^2 = 576$

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Now we can substitute back into our equation:

$48n + 576 = 5280$

$48n = 5280 - 576 = 4704$

$n = \dfrac{4704}{48} = 98$

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We need to find $1 + 2 + 3 + \ldots + 98$.

The formula for the sum of first $n$ natural numbers is:

$\text{Sum} = \dfrac{n(n+1)}{2}$

This formula works because if you write the sum forwards and backwards, then add them:

Forward: $1 + 2 + 3 + \ldots + n$

Backward: $n + (n-1) + (n-2) + \ldots + 1$

Adding: $(n+1) + (n+1) + (n+1) + \ldots + (n+1) = n(n+1)$

Since we added the sum to itself, the original sum is $\dfrac{n(n+1)}{2}$.

Therefore: $1 + 2 + 3 + \ldots + 98 = \dfrac{98 \times 99}{2} = \dfrac{9702}{2} = 4851$

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Answer: 4851

Key takeaways for similar problems:

1. Count terms carefully in arithmetic sequences
2. Factor out common parts to simplify
3. Remember: sum of first $k$ odd numbers = $k^2$
4. Remember: sum of first $n$ natural numbers = $\dfrac{n(n+1)}{2}$