Sequence Ak= 2a (k-1) - 3 if a1 =4 whats next 3 terms

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Sequence Ak= 2a (k-1) - 3 if a1 =4 whats next 3 terms

Mathematics
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K-1 is supposed to be low like a exponent but at bottom idk what its called?
index A2=2*4-3=5 A3=2*5-3=7 A4=2*7-3=11 A5=2*11-3=18 and so on
whats the star mean?

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multiply *
nevermind
thank you
An = 2(An-1) -3 An = 2^2 (An-2) - 2(3) - 3 An = 2^3 (An-3) - 2^2(3) - 2(3) - 3 An = 2^n(A0) - 3[x] x = 1 + 2 + 2^2 + 2^3 + ... + 2^n-1 2x = 2 + 2^2 + 2^3 + ... + 2^n-1 + 2^n 2x - x = 2^n - 1 x = (2^n -1)/2 An = 2^n(A0) - 3[(2^n -1)/2] perhaps?
might need to adjust for A1; tho, so lets check.... well, it dont match but i was close with it
is there a way I can write it with out typing so i can show you exactly how it looks
A1 = 4 A2= 2(4) - 3 = 5 A3= 2(5) - 3 = 7 A4= 2(7) - 3 = 11 is correct tho
write it without typing it.... prolly not
okay hold on ill use paint and up load it
or like this maybe you'll understand A = 2A -3 k (k-1)
but the k supposed to be up closer next to the a but set a little lower same for (k-1) Like when you do a LOG
do you understand now?
\[a _{k}\]
yesss :)
the solution is what I wrote, I cannot understand what amistre wrote :)
thank you do you have time for another
sure
Use mathematical induction 7+ 14+ 21+ 21 +28 +.....7N= 7N(N+1)/ 2
ok, do you now what induction is in maths?
7n(N+1) is all divided by 2 and not really
I know how to prove P(1) true but (k+1) is where im lost
induction is a technique to prove things has 3 steps 1. what you wrote: prove that it is true for N=1 2. Assume that it is true for any N 3. Prove that it is true for N+1
1. N=1 7=7*1(1+1)/2=7 we are happy :) 2. (called inductive step) Suppose this is true: 7+ 14+ 21+ 21 +28 +.....7N= 7N(N+1)/ 2 3. prove it for N+1 (we can use the inductive step) 7+ 14+ 21+ 21 +28 +.....7N+7(N+1)= this bit is 7+ 14+ 21+ 21 +28 +.....7N=7N(N+1)/ 2 so the whole is 7N(N+1)/ 2 +7(N+1)= 7(N+1)(N/2+1)=7(N+1)(N+2)/2 This completes the proof. (can you see why?)
and we are very happy now :)
thank you
I have more sorry its confusing to me
A ball bounces 400 ft every second after that its (3/4) how many times does it bounce til it completely stops
no prob
I figured its a geometric sequence
yep it is
so how do I solve?
I dont understand the problem completely
after that its (3/4) what does this mean?
so the common ratio is 3/4 400, 300... and so on
so 3/4 of 300 then the next and next.. but I need to find out how many times it bounces til it stops
It will never stop than
a number multiplied by 3/4 a lot of times will be a small number but will never reach 0 in maths we say it is tending to 0 as we tend to infinity
well until it hits the ground?
oh I am stupid, I think it asks for the use of this formula: a/(1-r)
here a=400 1-r=1/4 so the result is 1600
ohhh i seee :)

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