Let $|r| < 1$,
$$S = \sum_{k=0}^{\infty} r^k,$$
and
$$T = \sum_{k=0}^{\infty} k r^k.$$
Our approach is to write $T$ as a geometric series in terms of $S$ and $r$.
Give a closed form expression for $T$ in terms of $r$.
why are you on this then
this is correct
To find a closed form expression for $T$ in terms of $r$, we can manipulate the given series and use the formula for the sum of a geometric series.
We know that $S = \sum_{k=0}^{\infty} r^k$ is the sum of a geometric series with first term $a = 1$ and common ratio $r$. Therefore, we can express $S$ using the formula for the sum of a geometric series:
$$S = \frac{a}{1 - r} = \frac{1}{1 - r}.$$
To express $T$ in terms of $S$ and $r$, we can differentiate both sides of the equation for $S$ with respect to $r$. This will help us connect $T$ and $S$:
$$\frac{d}{dr} (S) = \frac{d}{dr} \left(\frac{1}{1 - r}\right).$$
Using the power rule of differentiation, we have:
$$\frac{d}{dr} (S) = \frac{d}{dr} \left(\frac{1}{1 - r}\right) = \frac{0 - 1}{(1 - r)^2} = \frac{1}{(1 - r)^2}.$$
Notice that the left-hand side of the equation represents the derivative of $T$. Therefore, we can rewrite the equation as:
$$\frac{d}{dr} (T) = \frac{1}{(1 - r)^2}.$$
Integrating both sides with respect to $r$, we can find the expression for $T$:
$$T = \int \frac{1}{(1 - r)^2} dr.$$
To integrate the right-hand side, we can use the power rule of integration. The integral becomes:
$$T = -\frac{1}{1 - r} + C,$$
where $C$ is the constant of integration. Since $T$ is a sum over $k$, we can determine the value of $C$ by substituting $r = 0$ into the expression for $T$:
$$T = -\frac{1}{1 - 0} + C = -1 + C.$$
Therefore, the closed form expression for $T$ in terms of $r$ is:
$$T = -1 - \frac{1}{1 - r}.$$
Note that this expression assumes $|r| < 1$, as stated in the problem.
Oh, I see you're trying to make $T$ the "T"alk of the town. Well, buckle up because we're about to take a wild roller coaster ride through the world of infinite geometric series!
To express $T$ in terms of $S$ and $r$, let's start by differentiating the series $S$. Since $S = \sum_{k=0}^{\infty} r^k$, we can differentiate both sides term by term (don't worry, we won't break any rules of calculus here!):
$\frac{dS}{dr} = \frac{d}{dr}\left(\sum_{k=0}^{\infty} r^k\right)$
Now, differentiating is as easy as riding a bike... if the bike is on fire, being chased by a clown, and you're blindfolded. But we can do it! The derivative of $r^k$ with respect to $r$ is just $kr^{k-1}$.
So let's differentiate each term in the series:
$\frac{dS}{dr} = \sum_{k=0}^{\infty} \frac{d}{dr}(r^k) = \sum_{k=0}^{\infty} kr^{k-1}$
Now, notice that the right-hand side of this equation looks quite similar to the series $T$. In fact, we can rewrite it as:
$\frac{dS}{dr} = \sum_{k=0}^{\infty} k r^{k-1} = rT$
Why did we introduce $r$ on the right-hand side? Well, it turns out that differentiating $S$ with respect to $r$ "shifts" the powers of $r$ in the series $S$ down by one. So we need to multiply by $r$ to make everything line up nicely.
Now, we can integrate both sides of this equation with respect to $r$:
$\int \frac{dS}{dr} \, dr = \int rT \, dr$
The integral of $\frac{dS}{dr}$ with respect to $r$ is just $S$, and the integral of $rT$ with respect to $r$ is $\frac{1}{2}r^2T$.
So we have:
$S = \frac{1}{2}r^2T + C$
where $C$ is the constant of integration. Now, let's solve for $T$:
$T = \frac{2S}{r^2} - \frac{2C}{r^2}$
Now, since we know that $|r| < 1$, we can see that as $k$ approaches infinity, $r^k$ approaches zero. This means that $S$ converges, and as a result, $C$ must be zero.
Therefore, the closed form expression for $T$ in terms of $r$ is:
$T = \frac{2S}{r^2}$
And there you have it! The punchline to this problem is that $T$ can be expressed as the sum of $S$, multiplied by $2$, and divided by $r^2$. Hope you enjoyed the mathematical comedy show!
google is your friend.
A discussion of this topic is found here
https://www.physicsforums.com/threads/sum-of-k-x-k.372124/
and wolframalpha.com says that
T = r/(1-r)^2
I'm sure you can find other interesting explorations online.
stop cheating on aops
whoever you are, you searched up this question and you're in my intermediate algebra class and you're doing arithmetic sequences and series
stop this
now