The sum is $$\sum_{k=5}^{\infty}\binom{k-1}{k-5}\frac{k^3}{2^{k}} $$

The first thing I thought of was the binomial coefficient. So I re-indexed it $$\sum_{j=4}^{\infty}\binom{j}{j-4}\frac{(j+1)^{3}}{2^{(j+1)}} $$ But then I realized that my interval wouldn't make sense if it was, leading me to think it's a taylor series manipulation. I'm not sure what I should be looking at. Any hints?

  • $\begingroup$ Velcome to our site! $\endgroup$ – kjetil b halvorsen Jul 30 '14 at 10:18
  • 1
    $\begingroup$ For starters, note that $\binom{k-1}{k-5} = \binom{k-1}{4}$ and $\binom{j}{j-4}=\binom{j}{4}$. $\endgroup$ – Dilip Sarwate Jul 30 '14 at 10:56


$$f(x; n, r, s) = \sum_{k=0}^\infty \binom{k+s}{k} (k+r)^n x^{k+r}$$

for nonnegative integers $n, r, s$. By means of the differential operator $D: x^j\to jx^{j-1}$ and the formulae

$$\binom{k+s}{k}x^{k+r} =\left(\frac{1}{s!}(k+s)(k+s-1)\cdots(k+1)x^k\right)x^r = \frac{1}{s!}x^rD^s x^{k+s};$$

$$(k+r)^n x^{k+r} = (xD)^n x^{k+r},$$

this is readily seen to equal

$$f(x; n, r, s) = (xD)^n\left(\frac{1}{s!} x^r D^s\left( x^s\sum_{k=0}^\infty x^k\right)\right) = \frac{1}{s!}(xD)^n \left(x^r D^s \frac{x^s}{1-x}\right)$$

provided $|x| \lt 1$ (for which all series that occur in the calculation converge absolutely, permitting interchanging the summation with the differential operators).

Because (re-indexing to start from zero)

$$\sum_{k=5}^{\infty}\binom{k-1}{k-5}\frac{k^3}{2^{k}} = \sum_{k=0}^\infty\binom{k+4}{k}(k+5)^3\left(\frac{1}{2}\right)^{k+5},$$

you seek the value of $f(x; 3, 5, 4) = 5 x^5 (25 + 16 x + x^2)/(1-x)^8$ at $x=1/2$, where it equals $1330$.


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