What is the real number a that satisfies the equation a+2a^2+3a^3+4a^4+...=30?

[anemone]

Find the real number $a$ such that $a+2a^2+3a^3+4a^4+\cdots=30$.

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[anemone]

Re: Problem of the week #103 -March 17th, 2014

Congratulations to the following members for their correct solutions::)

1. MarkFL
2. magneto
3. kaliprasad
4. lfdahl
5. Pranav

Solution from magneto:

Spoiler

Consider the sum $S = \sum_{k=0}^\infty (k+1)a^k$. The series we are interested in (LHS) is therefore:
$aS = a\sum_{k=0}^\infty (k+1)a^k$. However, we know that $S - aS = \sum_{k=0}^\infty a^k = \frac{1}{1-a}$.
So, we have that $S(1-a) = \frac{1}{1-a}$, or that $S = \frac{1}{(1-a)^2}$.

The LHS of the equation is $aS = \frac{a}{(1-a)^2}$. We can solve the equation $a = 30(1 - 2a + a^2)$, or
$30a^2 - 61a + 30 = 0$. Using the quadratic formula, we have that $a = \frac{5}{6}$ or $a = \frac{6}{5}$.
We will reject the answer where $|a| > 1$ as it will cause the series to diverge.

Hence, $a = \frac{5}{6}$.

Solution from lfdahl:

Spoiler

\[Let \;\;\; S = a+2a^2+3a^3+4a^4+ ...=30 \\\\
Then \;\;\;
aS = a^2+2a^3+3a^4+4a^5+ \; ... \\\\
\Rightarrow aS + a + a^2+a^3+a^4+ \; ...=S \\\\
\Rightarrow aS+\frac{a}{1-a} = S, \;\;\; |a| < 1\\\\
\Rightarrow a^2-(2+S^{-1})a+1=0 \\\\
\Rightarrow a=\frac{1}{2}(2+S^{-1}\pm 11S^{-1})\Rightarrow a = \frac{5}{6}\]