bumblebee123
Junior Member
- Joined
- Jan 3, 2018
- Messages
- 200
Pka is starting with the exotic "representative formula" with the scary
"sum symbol" (the one that scares most students!).
The formula you use is correct (same as Pka's):
Sum of 1st n terms = n/2 [2a + ( n - 1 )d ]
Usually shown this way: n[2a + d(n - 1)] / 2
So we have for n terms:
n[2a + d(n - 1)] / 2
= n(2a + dn - d) / 2 ; substitute d = 2a
= n(2a + 2an - 2a) / 2
= (2an + 2an^2 - 2an) / 2
= 2an^2 / 2
= an^2
Now for 2n terms; we start similarly with:
n[2a + d(n - 1)] / 2
= n(2a + dn - d) / 2 ; substitute d = 2a and n = 2n:
= 2n(2a + 4an - 2a) / 2
= n(2a + 4an - 2a)
= 2an + 4an^2 - 2an
= 4an^2
It always helps if you make up a simple case; as example:
3, 9, 15 ,21, 27, 33
a = 3
d = 2a = 6
n = 3 (so 2n = 6)
3 + 9 + 15 = 27
3 + 9 + 15 + 21 + 27 + 33 = 108
108 / 27 = 4
would this work if I used my formula? ( as I'm not used to the other one )
a1 = a d = 2a n = N
sN = N/2 [ 2a + ( N -1 ) 2a ] = N/2 ( 2a + 2aN - 2a ) = N/2 ( 2aN ) = N ( aN ) = aN^2
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