Evaluate the sum of (i² - 1) for i from 2 to 5.

Step-by-step solution

1. 1

   Read what the sigma notation says. The index $i$ runs through the integers $2,3,4,5$ - four terms, not five, because both limits are inclusive. For each one, substitute into $i^2-1$ and add the results.

2. 2

   Evaluate each term. $i=2$: $4-1=3$. $i=3$: $9-1=8$. $i=4$: $16-1=15$. $i=5$: $25-1=24$. Note each term is $i^2-1=(i-1)(i+1)$, so the values are $1\cdot3$, $2\cdot4$, $3\cdot5$, $4\cdot6$.

3. 3

   Add the four terms. $3+8+15+24 = 50.$ Pairing helpfully: $3+24=27$ and $8+15=23$, and $27+23=50$.

4. 4

   Verify with the closed forms. Split the sum: $\sum_{i=2}^{5} i^2 - \sum_{i=2}^{5} 1$. The first is $\tfrac{5\cdot6\cdot11}{6}-1 = 55-1 = 54$ (sum of squares to $5$, minus the $i=1$ term), and the second is simply $4$. Then $54-4=50$, matching.

5. 5

   Watch the two classic traps. Starting the index at $1$ instead of $2$ would add $1^2-1=0$ and coincidentally not change the answer here, but subtracting $1$ only once at the end (giving $54-1=53$) is a genuine error - the $-1$ appears inside the summand, so it is subtracted four times.