Here are the first five terms of an arithmetic sequence. 7
13
19
25
31
Prove that the difference between the squares of any two terms of the sequence is always
a multiple of 24.

9514 1404 393

  the difference of squares is 24(3n+2), so a multiple of 24

Step-by-step explanation:

The sequence has a first term of 7 and a common difference of 6, so the n-th term is ...

  a[n] = a1 +d(n -1)

  a[n] = 7 + 6(n -1)

  a[n] = 6n +1

Then the (n+1)-th term is ...

  a[n+1] = 6(n+1) +1 = 6n +7

The difference of the squares of these terms is ...

  a[n+1]^2 -a[n]^2 = (6n+7)^2 -(6n+1)^2

  = (36n^2 +84n +49) -(36n^2 +12n +1)

  = 72n +48

  = 24(3n +2) . . . . . always a multiple of 24

The difference of squares of terms n and n+1 is 24(3n+2), a multiple of 24.

Check

The difference between squares for n=1 and n=2 is 13^2 -7^2 = 169 -49 = 120. The formula tells us the difference is 24(3·1 +2) = 24·5 = 120, in agreement.

correct answer:

i^31 = -i

i think its doubled. let me know

step-by-step explanation: