Find the sum of the infinite series 1/3+4/9+16/27+64/81+.... if it exists.
1 answer:
9/27, 12/27, 16/27
So this is a geometric sequence as each term is 4/3 the previous term.
Since the common ration is greater than one the sum of the series diverges, it does not exist. (The sum just keeps getting larger and larger)
For a geometric series to have a sum r^2<1
So that the normal sum....
s(n)=a(1-r^n)/(1-r) becomes if r^2<1
s=a/(1-r)
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27 * 35
= 20*35 + 7 * 35
= 700 + 245
= 945
57.10
70.9
these are only the answers for the first two problems
- y^2+9y+20
- y^2+10y+21
- y^2+2y+3
- Step-by-step explanation:
- (y+5(y+4) =y(y+4)+5(y+4) =y^2+4y+5y+20 =y^2+9y+20
- (y+3(y+7) =y(y+7)+3(y+7) =y^2+7y+3y+21 =y^2+10y+21
- (y+3)(y-1)
=y(y-1)+3(y-1)
=y^2-y+3y-3
= y^2+2y+3
Answer:
Well no cheating not from me
Step-by-step explanation:
Answer:
4.55
Step-by-step explanation: