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10 votes

$a+a r+a r^(2)+\ldots \infty=15$$a^(2)+(a r)^(2)+\left(a r^(2)\right)^(2)+\ldots \infty=150$. Find $a r^(3)+a r^(4)+a r^(6)+\ldots \infty$

Options:

(a) $(1)/(2)$\\(b) $(2)/(5)$

User Kevin Meyer
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1 Answer

10 votes
10 votes

Let


S_n = \displaystyle \sum_(k=0)^n r^k = 1 + r + r^2 + \cdots + r^n

where we assume |r| < 1. Multiplying on both sides by r gives


r S_n = \displaystyle \sum_(k=0)^n r^(k+1) = r + r^2 + r^3 + \cdots + r^(n+1)

and subtracting this from
S_n gives


(1 - r) S_n = 1 - r^(n+1) \implies S_n = (1 - r^(n+1))/(1 - r)

As n → ∞, the exponential term will converge to 0, and the partial sums
S_n will converge to


\displaystyle \lim_(n\to\infty) S_n = \frac1{1-r}

Now, we're given


a + ar + ar^2 + \cdots = 15 \implies 1 + r + r^2 + \cdots = \frac{15}a


a^2 + a^2r^2 + a^2r^4 + \cdots = 150 \implies 1 + r^2 + r^4 + \cdots = (150)/(a^2)

We must have |r| < 1 since both sums converge, so


\frac{15}a = \frac1{1-r}


(150)/(a^2) = \frac1{1-r^2}

Solving for r by substitution, we have


\frac{15}a = \frac1{1-r} \implies a = 15(1-r)


(150)/(225(1-r)^2) = \frac1{1-r^2}

Recalling the difference of squares identity, we have


\frac2{3(1-r)^2} = \frac1{(1-r)(1+r)}

We've already confirmed r ≠ 1, so we can simplify this to


\frac2{3(1-r)} = \frac1{1+r} \implies (1-r)/(1+r) = \frac23 \implies r = \frac15

It follows that


\frac a{1-r} = \frac a{1-\frac15} = 15 \implies a = 12

and so the sum we want is


ar^3 + ar^4 + ar^6 + \cdots = 15 - a - ar - ar^2 = \boxed{\frac3{25}}

which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?

User Housefly
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