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Question 1209814: Let r be a real number such that |r| < 1. Express
\sum_{n = 0}^{\infty} n*r^n*(n + 1)*(n + 2)
in terms of r.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let $S = \sum_{n=0}^{\infty} n(n+1)(n+2)r^n$.
Consider the geometric series $\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}$ for $|r| < 1$.
Differentiating with respect to $r$, we have:
$\sum_{n=1}^{\infty} nr^{n-1} = \frac{1}{(1-r)^2}$.
Multiplying by $r$, we get:
$\sum_{n=1}^{\infty} nr^n = \frac{r}{(1-r)^2}$.
Differentiating again with respect to $r$:
$\sum_{n=1}^{\infty} n^2 r^{n-1} = \frac{(1-r)^2 + 2r(1-r)}{(1-r)^4} = \frac{(1-r)(1-r+2r)}{(1-r)^4} = \frac{1+r}{(1-r)^3}$.
Multiplying by $r$:
$\sum_{n=1}^{\infty} n^2 r^n = \frac{r(1+r)}{(1-r)^3}$.
Differentiating again with respect to $r$:
$\sum_{n=1}^{\infty} n^3 r^{n-1} = \frac{(1-r)^3(1+2r) + 3r(1+r)(1-r)^2}{(1-r)^6} = \frac{(1-r)(1+2r) + 3r(1+r)}{(1-r)^4} = \frac{1+2r-r-2r^2+3r+3r^2}{(1-r)^4} = \frac{r^2+4r+1}{(1-r)^4}$.
Multiplying by $r$:
$\sum_{n=1}^{\infty} n^3 r^n = \frac{r(r^2+4r+1)}{(1-r)^4}$.
Now we want to find $\sum_{n=0}^{\infty} n(n+1)(n+2)r^n$.
Note that $n(n+1)(n+2) = n(n^2+3n+2) = n^3 + 3n^2 + 2n$.
Therefore,
$S = \sum_{n=0}^{\infty} (n^3 + 3n^2 + 2n)r^n = \sum_{n=0}^{\infty} n^3 r^n + 3\sum_{n=0}^{\infty} n^2 r^n + 2\sum_{n=0}^{\infty} nr^n$.
Using the results above, we have:
$S = \frac{r(r^2+4r+1)}{(1-r)^4} + 3\frac{r(1+r)}{(1-r)^3} + 2\frac{r}{(1-r)^2}$.
$S = \frac{r(r^2+4r+1) + 3r(1+r)(1-r) + 2r(1-r)^2}{(1-r)^4}$.
$S = \frac{r(r^2+4r+1) + 3r(1-r^2) + 2r(1-2r+r^2)}{(1-r)^4}$.
$S = \frac{r^3+4r^2+r + 3r-3r^3 + 2r-4r^2+2r^3}{(1-r)^4}$.
$S = \frac{r^3+4r^2+r+3r-3r^3+2r-4r^2+2r^3}{(1-r)^4}$.
$S = \frac{(1-3+2)r^3 + (4-4)r^2 + (1+3+2)r}{(1-r)^4}$.
$S = \frac{6r}{(1-r)^4}$.
Final Answer: The final answer is $\boxed{\frac{6r}{(1-r)^4}}$
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