Solving the Alternating Squares Series

The problem asks for the value of the series: $ S = 1^2-2^2 + 3^2-4^2 + 5^2+...+17^2-18^2 + 19^2 $ We can group the terms in pairs and use the difference of squares formula, $a^2 - b^2 = (a-b)(a+b)$.

Grouping Terms using Difference of Squares

Group the series into pairs: $ S = (1^2-2^2) + (3^2-4^2) + (5^2-6^2) + ... + (17^2-18^2) + 19^2 $ Apply the difference of squares formula to each pair:

• $1^2 - 2^2 = (1-2)(1+2) = (-1)(3) = -3$
• $3^2 - 4^2 = (3-4)(3+4) = (-1)(7) = -7$
• $5^2 - 6^2 = (5-6)(5+6) = (-1)(11) = -11$
• ...
• $17^2 - 18^2 = (17-18)(17+18) = (-1)(35) = -35$

The series now becomes: $ S = -3 + (-7) + (-11) + ... + (-35) + 19^2 $

Summing the Arithmetic Progression

The terms $-3, -7, -11, ..., -35$ form an arithmetic progression (AP).

• First term ($a$) = -3
• Common difference ($d$) = -7 - (-3) = -4
• Last term ($l$) = -35

To find the number of terms ($n$) in this AP, we consider the first term of each pair (1, 3, 5, ..., 17). The number of terms is $n = \frac{\text{Last term} - \text{First term}}{\text{Common difference}} + 1 = \frac{17-1}{2} + 1 = \frac{16}{2} + 1 = 8 + 1 = 9$.

Use the sum formula for an AP: $S_n = \frac{n}{2}(a+l)$. $ S_9 = \frac{9}{2}(-3 + (-35)) = \frac{9}{2}(-38) = 9 \times (-19) = -171 $

Calculating the Final Value

Now, substitute the sum of the AP back into the series expression:

Thus, the value of the series is 190.