The distance formula, d= √ (x2-x1)^2+(y2-y1)^2, is the key element in the proof of which of the following?
1 answer:
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Part (i)
I'm going to use the notation T(n) instead of
To find the first term, we plug in n = 1
T(n) = 2 - 3n
T(1) = 2 - 3(1)
T(1) = -1
The first term is -1
Repeat for n = 2 to find the second term
T(n) = 2 - 3n
T(2) = 2 - 3(2)
T(2) = -4
The second term is -4
<h3>Answers: -1, -4</h3>==============================================
Part (ii)
Plug in T(n) = -61 and solve for n
T(n) = 2 - 3n
-61 = 2 - 3n
-61-2 = -3n
-63 = -3n
-3n = -63
n = -63/(-3)
n = 21
Note that plugging in n = 21 leads to T(21) = -61, similar to how we computed the items back in part (i).
<h3>Answer: 21st term</h3>===============================================
Part (iii)
We're given that T(n) = 2 - 3n
Let's compute T(2n). We do so by replacing every copy of n with 2n like so
T(n) = 2 - 3n
T(2n) = 2 - 3(2n)
T(2n) = 2 - 6n
Now subtract T(2n) from T(n)
T(n) - T(2n) = (2-3n) - (2-6n)
T(n) - T(2n) = 2-3n - 2+6n
T(n) - T(2n) = 3n
Then set this equal to 24 and solve for n
T(n) - T(2n) = 24
3n = 24
n = 24/3
n = 8
This means 2n = 2*8 = 16. So subtracting T(8) - T(16) will get us 24.
<h3>Answer: 8</h3>