#How can i simplify expressions like this?
47 messages · Page 1 of 1 (latest)
full expression:
(1+y) + (2+y)y + (3+y)y^2 ... ( (c-1) + y )*y^(c-2)
How can i simplify expressions like this?
think of teach term in this series as (1 + y), (2 + y)y, (3 + y)yˆ2...
now lets get this in a form thats easier to work with
we can rewrite (2 + y)y as y + (1 + y)y, and (3 + y)yˆ2 as 2yˆ2 + (1 + y)y^2
so now our terms look like
1 + y
- y + (1 + y)y
- 2y^2 + (1 + y)y^2
- 3y^3 + (1 + y)y^3
- ...
so what can we do with this now to simplify it
ohhhh
lemme try
(1+y)
y + (1 + y)y
2y^2 + (1+y)^2
...
(c-2)y^(c-2) + (1+y)^(c-2)
this right?
@sinful lava
(c-2)y^(c-2) + (1+y)*y^(c-2)
y^(c-2) ( (c-2) + (1 +y) )
y^(c-2) ( c-1 + y)
yes this is right
1 + y
y + (1 + y)y
2y^2 + (1 + y)y^2
3y^3 + (1 + y)y^3
...
(c-2)y^(c-2) + (1+y)*y^(c-2)
1 + y
y + (1 + y)y
2y^2 + (1 + y)y^2
3y^3 + (1 + y)y^3
...
(c-2)y^(c-2) + (1+y)*y^(c-2)
lets separate this
y + 2y^2 + 3y^3 ... + (c-2)y^(c-2) +
(1+y) + (1+y)y + (1+y)y^2 ... (1+y)^(c-2)
just use the formule on the below
s = (1+y) + (1+y)y + (1+y)y^2 ... (1+y)^(c-2)
s = (1+y) * ( y^(c-1) -1)/(y-1)
now
we got
y + 2y^2 + 3y^3 ... + (c-2)y^(c-2) + (1+y) * ( y^(c-1) -1)/(y-1)
how can i factor the other sum?
yesssss you gthe goat
?
like
theres no common multiples ( only y but i cant do anything with it)
z = y + 2y^2 + 3y^3 + 4y^4 ....
= y
- y^2 + 2y^3 + 3y^4...
- y^2 + y^3 + y^4...
= y
- y(y + 2y^2 + 3y^3...)
- y^2 + y^3 + y^4...
= y
- y(z)
- y^2 + y^3 + y^4...
z = y + yz + (y^2 + y^3 + y^4...)
oh
z is the solution to the other sum
alr
this is for the infinite form though
i have to go now but
hopefully this is enough for you
to do the finite form
cya man
and i pray that this is not on your test
what im trying to do is proof compound interest formula
u iwll close?
sure
just incase you have other questions for anyone else to answer