If `S_n=1^2-2^2+3^2-4^2+5^2-6^2+ ,t h e n` `S_(40)=-820` b. `S_(2n)> S_(2n+2)` c. `S_(51)=1326` d. `S_(2n+1)> S_(2n-1)`

Options

A. `S_(40)=-820`
B. `S_(2n) gt S_(2n+2)`
C. `S_(51)=1326`
D. `S_(2n +1) gt S_(2n-1)`

Explanation

To solve the problem, we will analyze the series \( S_n = 1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + \ldots \) and verify the given statements one by one.

### Step 1: Finding \( S_n \) for even \( n \)

For even \( n \), we can pair the terms:
\[ S_n = (1^2 - 2^2) + (3^2 - 4^2) + (5^2 - 6^2) + \ldots + ((n-1)^2 - n^2) \]
Each pair can be simplified:
\[ k^2 - (k+1)^2 = k^2 - (k^2 + 2k + 1) = -2k - 1 \]
Thus, we can express \( S_n \) as:
\[ S_n = -\left(2(1 + 3 + 5 + \ldots + (n-1)) + \frac{n}{2}\right) \]
The sum of the first \( m \) odd numbers is \( m^2 \), where \( m = \frac{n}{2} \):
\[ S_n = -\left(2 \left(\frac{n}{2}\right)^2 + \frac{n}{2}\right) = -\left(\frac{n^2}{2} + \frac{n}{2}\right) = -\frac{n(n+1)}{4} \]

### Step 2: Finding \( S_n \) for odd \( n \)

For odd \( n \):
\[ S_n = (1^2 - 2^2) + (3^2 - 4^2) + (5^2 - 6^2) + \ldots + (n^2) \]
The last term is \( n^2 \), and we can pair the terms similarly:
\[ S_n = -\left(2(1 + 3 + 5 + \ldots + (n-2)) + n^2\right) \]
The sum of the first \( m \) odd numbers is again \( m^2 \), where \( m = \frac{n-1}{2} \):
\[ S_n = -\left(2 \left(\frac{n-1}{2}\right)^2 + n^2\right) = -\left(\frac{(n-1)^2}{2} + n^2\right) \]
Simplifying gives:
\[ S_n = -\frac{n^2 + n^2 - 2n + 1}{2} = -\frac{2n^2 - 2n + 1}{2} = -n(n + 1)/2 \]

### Step 3: Calculate \( S_{40} \)

Since \( 40 \) is even:
\[ S_{40} = -\frac{40 \times 41}{4} = -410 \]

### Step 4: Verify the options

a. \( S_{40} = -820 \) is **not true** since we calculated \( S_{40} = -410 \).

b. \( S_{2n} > S_{2n+2} \):
\[ S_{2n} = -\frac{2n(2n + 1)}{4} = -\frac{n(2n + 1)}{2} \]
\[ S_{2n+2} = -\frac{(2n+2)(2n+3)}{4} \]
Since both are negative, \( S_{2n} \) is greater than \( S_{2n+2} \).

c. \( S_{51} = 1326 \):
\[ S_{51} = -\frac{51 \times 52}{4} = -663 \text{ (not true)} \]

d. \( S_{2n+1} > S_{2n-1} \):
\[ S_{2n+1} = -\frac{(2n+1)(2n+2)}{4} \]
\[ S_{2n-1} = -\frac{(2n-1)(2n)}{4} \]
Both are negative, but \( S_{2n+1} \) is less negative than \( S_{2n-1} \).

### Conclusion

Thus, the only true statements are:
- b. \( S_{2n} > S_{2n+2} \)