#Arithmetic Sequences
grave oracle
a=2t+1
nth term=14t-5
d=3
S(n)=p(qt-1)^r
Sn=n/2(2a+d(n-1))
Sn=n/2(4t+2+3(n-1))
2t+1 +3(n-1)=14t-5
3(n-1)=12t-6
t=1/4(n+1) - but this is not an identity as n is not constant
12t is a multiple of 3.
2t+1+3(n-1)=14t-5
12t-3=3n
nth term here=4t-1
2t+1+3(4t-1-1)=14t-5
Assuming n is is the same n as that of the nth term of line 2 of the question:
S4t-1=(2t-0.5)(4t+2+12t-6)=(2t-0.5)(16t-4)=32t^2-16t+2
32(t^2-1/2)+2
32(t-1/4)^2 +2
(t-1/4)^2=t^2+1/16-t/2. t^2-1/2=t^2+1/16-t/2 -9/16 +t/2
32(t-1/4)^2 +23/16+t/2
How do I get rid of the spaced out bit on the end of the line above.
coefficient of t is 1 not q... q is probably >1
im pretty sure they want you to complete the square on the sum equation. but without unknown constants (besides t) because u r meant to have solved for the others.
i think that i did that.
grave oracle
.
chilly dew
When you first get the quadratic you can factor out 2 to get 2(16t² - 8t + 1) which is a perfect square that can factorise to give 2(4t - 1)²
chilly dew
a=2t+1
nth term=14t-5
d=3
S(n)=p(qt-1)^r
Sn=n/2(2a+d(n-1))
Sn=n/2(4t+2+3(n-1))...
Or if you didn’t see it was a perfect square then you can complete the square to get
32(t - ¹/₄)²
But since we want -1 and not -1/4 we can rewrite this as
32[¹/₄(4t - 1)]²
When you distribute the power you get 32[¹/₁₆(4t - 1)²] which is 2(4t - 1)²
grave oracle
When you first get the quadratic you can factor out 2 to get 2(16t² - 8t + 1) wh...
Thanks
am i just supposed to be able to spot that