Question

\[\sum_{r=1}^n r^3 \binom{n}{r}=n^2 (n+3)2^{n-3}\]Prove.

Answer 1

@danica518 @dan815

Answer 2

hello bby

Answer 3

hi sweetie

Answer 4

k lets start with k=n+3

Answer 5

After you're done, I'll share a really nice solution with you. <3

Answer 6

oh then show me lol.. im eating

Answer 7

Now express\[r^3 = r(r-1)(r-2) + 3r(r-1) + r\]And we're done.

Answer 8

ohh haha nice!!

Answer 9

Whoopers, I forgot all those binomial coefficients there.

Answer 10

Alright.\[(1+x)^n = \sum_{r =0}^{n}\binom{n}{r}x^{r}\]Differentiate.\[n(1+x)^{n-1}=\sum_{r=0}^{n}r\binom{n}{r} x^{r-1}\]DIFFERENTIATE.\[n(n-1) (1+x)^{n-2}=\sum_{r=0}^{n}r(r-1)\binom{n}r x^{r-2}\]DIFFFFEEEEERENTIATEEE!!!\[n(n-1)(n-2) (1+x)^{n-3}=\sum_{r=0}^{n}r(r-1)(r-2) \binom{n}rx^{r-3}\]

Answer 11

\[n(n-1)(n-2)\cdot 2^{n-3} + 3n(n-1)\cdot2^{n-2}+n\cdot2^{n-1}\]\[= (n^3 - 3n^2 + 2n + 6n^2 - 6n + 4n)\cdot 2^{n-3}\]\[= (n^3 + 3n^2)\cdot 2^{n-3}\]Just to finish the deal.

Answer 12

kewl

Answer 13

Here I tried to generalize it using:

\[\frac{d^k}{dx^k} (x^r) = \frac{r!}{(r-k)!}x^{r-k}\]
Starting here:
\[(1+x)^n=\sum_{r=0}^n \binom{n}{r} x^r\]
\[\frac{n!}{(n-k)!}(1+x)^{n-k}=\sum_{r=0}^n \frac{n!}{r! (n-r)!}
\frac{r!}{(r-k)!}x^{r-k}\]

Some monkeying around:


\[(1+x)^{n-k}=\sum_{r=0}^n \frac{(n-k)!}{(n-r)!(r-k)!}x^{r-k}\]
More playing:


\[(1+x)^n \left(1+\frac{1}{x} \right)^k=\sum_{r=0}^n \binom{n-k}{r-k}x^r\]
Uhhh I don't know if I got anywhere.

Answer 14

whoops I think I messed up that fraction on the left there in the final step, it should be \[\left(\frac{x}{x+1} \right)^k\]

Answer 15

So that's equal to\[x^k(1+x)^{n-k}\]The right-hand-side kinda makes sense that way. The \(x^k\) causes the coefficients to "shift" in a way.

Answer 16

Yeah I'm thinking we might be able to get a recurrence relation out of this or maybe not idk. I gotta go study so I'll check in here now and then.

Answer 17

Aight, good luck with your chemistry.