Calculate $\sum_1^{30}r^3 {30 \choose r}$

Question

How do I calculate $\sum_{r=1}^{30}r^3 {30 \choose r}$.

My approach is lengthy $(1+x)^n$ Differentiate it then multiply by x

Again differentiate it then multiply by x.

Repeat this process again we get the required summation. Is there any other shorter method.

It is lengthy and cumbersome.

Answer 1

Combinatorial Proof of general case:

$\displaystyle\sum_{r=0}^{r=n}r^3.^nC_{r}=n^2(n+3).2^{n-2}$

LHS is say no. of ways to form student council of size $r$ out of $n$ students in the class.Then from this council elect House captain,Head boy & Cultural Captain(these all can be same person also) .

for RHS: select one student in $\ n$ ways and give him all three positions then rest can be part of council in $2^{n-1}$ ways giving total no. of such councils as $n.2^{n-1}$

OR

select $2$ students in $n.(n-1)$ ways for occupying those $3$ designations in $^3C_2=3$ways and rest members in $2^{n-2}$ ways to give no. of such councils as $3n(n-1)2^{n-2}$

OR

select $3$ students in $n.(n-1)(n-2)$ ways for occupying those $3$ designations in $^3C_3=1$way and rest members in $2^{n-3}$ ways to give no. of such councils as $n(n-1)(n-2)2^{n-3}$

now add all three cases =$[n(n-1)(n-2)2^{n-3}+3n(n-1)2^{n-2}+n.2^{n-1}]$

$=2^{n-3}[n(n^2-3n+2)+4n+6n^2-6n]$

$=2^{n-3}[n^3+3n^2]$

$=2^{n-3}n^2[3+n]$ ways to form council.

for your question put $n=30$

so, answer will be

$2^{27}\times 900 \times 33=29700\times 2^{27}$

Answer 2

Hint

Could you prove that $$\sum_{r=1}^{p}r^3 {p \choose r}=2^{p-3} p^2 (p+3)$$

Answer 3

$$\sum_{r=0}^n {n \choose r} x^r=(1+x)^n$$ Differetiate w.r.t. $x$ to get $$\sum_{r=0}^{n} r{n \choose r} x^r =nx(1+x)^{n-1}$$ Again differetiate w.r.t. $x$ to get $$\sum_{r=0}^n r^2 {n \choose r} x^{r} =(n^2x^2+nx)(1+x)^{n-2}.$$ Once more dofferentiating w.r.t. x, we get $$\sum_{r=0}^n r^3 {n \choose r} x^r=nx(n^2x^2+3nx-x+1)(1+x)^{n-3}$$ Now let us put $x=1$ here to get $$\sum_{r=0}^n r^3 {n \choose r} x^r=n(n^2+3n)2^{n-3}.$$ Finally by putting $n=30$, we get $$\sum_{r=0}^{30} r^3 {30 \choose r}=29700 \times 2^{27}$$

Answer 4

Write $$r^3=r(r-1)(r-2)+ar(r-1)+br\ \ \ \ (1)$$ where $a,b$ are arbitrary constants

so that $$r^3\binom nr=n(n-1)(n-2)\binom{n-3}{r-3}+an(n-1)\binom{n-2}{r-2}+bn\binom{n-1}{r-1}$$ as $$r(r-1)\cdot(r-s+1)\binom nr$$ $$=n(n-1)\cdots(n-s+1)\cdot\dfrac{(n-s)!}{(n-s-(r-s))!(r-s)!}$$ $$=n(n-1)\cdots(n-s+1)\binom{n-s}{r-s}$$ for $r>s\ge1$

Now use $$\sum_{r=0}^m\binom mr=(1+1)^m$$

Finally set $r=1,2$ in $(1)$ one by one to find $a,b$

Answer 5

For a combinatorial proof, consider the number of nonempty committees selected from 30 people, with a president, secretary, and treasurer, where these roles are not necessarily distinct people.