Alternating Sum of Squares
Define the alternating sum of squares:
$$S_n = (2n)^2 - (2n-1)^2 + (2n-2)^2 - (2n-3)^2 + \cdots + 2^2 - 1^2$$
Find a closed-form expression for $S_n$, and compute $S_{50}$.
Hints
- The alternating signs suggest pairing consecutive terms. What happens when you group $(2i)^2 - (2i-1)^2$ together?
- Use the identity $a^2 - b^2 = (a+b)(a-b)$ on each pair. What does $(2i)^2 - (2i-1)^2$ simplify to?
- After simplifying each pair to $4i - 1$, you need $\sum_{i=1}^{n}(4i - 1)$. Use the formula for $\sum_{i=1}^{n} i = n(n+1)/2$.
Worked Solution
How to Think About It: This is a telescoping/pairing problem. The terms alternate in sign, which screams "pair them up." Each pair is a difference of consecutive squares, and there is a clean factorization for $a^2 - b^2 = (a+b)(a-b)$. Once you pair and simplify, you will get a sum of a simple arithmetic expression.
Quick Estimate: For $n = 50$, the sum starts with $100^2 - 99^2 = 199$ and ends with $2^2 - 1^2 = 3$. There are 50 pairs. The average pair value is roughly $(199 + 3)/2 = 101$, so the sum is approximately $50 \times 101 = 5050$. Let's see if the exact answer matches.
Approach: Group the $2n$ terms into $n$ consecutive pairs, apply the difference-of-squares identity, and sum.
Formal Solution:
Group the terms into pairs of the form $(2i)^2 - (2i-1)^2$ for $i = 1, 2, \ldots, n$:
$$S_n = \sum_{i=1}^{n} \left[(2i)^2 - (2i-1)^2\right]$$
Apply the difference-of-squares factorization:
$$(2i)^2 - (2i-1)^2 = (2i + 2i - 1)(2i - 2i + 1) = (4i - 1)(1) = 4i - 1$$
So:
$$S_n = \sum_{i=1}^{n} (4i - 1) = 4 \sum_{i=1}^{n} i - \sum_{i=1}^{n} 1 = 4 \cdot \frac{n(n+1)}{2} - n = 2n(n+1) - n = n(2n + 1)$$
For $n = 50$:
$$S_{50} = 50 \times 101 = 5050$$
Answer: $S_n = n(2n+1)$, and $S_{50} = 5050$.
Intuition
Whenever you see an alternating sum, the first move is to pair adjacent terms. For sums of squares specifically, the difference-of-squares identity $a^2 - b^2 = (a+b)(a-b)$ turns a complicated expression into a trivial arithmetic sum. This is a standard trick that appears frequently in mental math competitions and quant interviews.
Notice that $S_{50} = 5050 = \sum_{i=1}^{100} i$. This is not a coincidence: $n(2n+1) = \sum_{i=1}^{2n} i$ when... actually it is not quite the same (the triangular number formula gives $2n(2n+1)/2 = n(2n+1)$, so they match exactly!). The alternating sum of the first $2n$ squares equals the sum of the first $2n$ integers. That is a nice identity worth remembering.