See you Need to round it so it is c=-4
I’m not exactly sure but I think II and IV.
B and C are absurd; if a series converges, it must have a sum, but if a series diverges, it cannot have a sum.
Now, notice that

That is, we can write the sum more compactly as

The series is geometric with common ratio

, so the series converges (and thereby has a sum), so the answer is D.
Answer:
here down is the answer
(2y/5y)^3+4y^2+6y
8y^3/125y^3 +16y^2+6y
8/125+16y^2+6y
8/125+16y^2+6y=0
16y^2+6y+8/125=0
it is in the form of quadratic equation
formula :
x=(-b+-square root of b^2-4ac) /2a
here a=16,b=6&c=8/125
x=(-b+-square root of b^2-4ac)/2a
x=(-6+-square root of 36-4.09)/2a here 4.09 came by calculation
x=(-6+-31.91)/2
so
x=(-6+31.91)/2 first (+);
x=25.91/2
x=12.955;
x=(-6-31.91)/2 then(-);
x=-37.91/2
x=-18.955 ;
ans is = 12.955&-18.955;
Answer:
i really dont know and do you know how i really dont know
Step-by-step explanation:
it bc i dont literally