0
$\begingroup$

Evaluate the following;

$$\sum_{r=0}^{50} (r+1) ^{1000-r}C_{50-r}$$

Using $^{n}C_{r}=^{n}C_{n-r}$ we get $\sum_{r=0}^{50} (r+1) ^{1000-r}C_{950}$

but I am not getting how to solve $\sum_{r=0}^{50} r \cdot \hspace{0.5 mm} ^{1000-r}C_{950}$

  • 0
    do you mean $\binom{1000-r}{50-r}$? Also, you can do $\binom{n}{k}$ via `\binom{n}{k}`. – 2017-01-15
  • 0
    @Mark Yeah I mean $\binom {100-r}{50-r}$ – 2017-01-15
  • 0
    If it is $100$ and not $1000$, them please edit the question. – 2017-01-15
  • 0
    @SchrodingersCat It is 1000 only – 2017-01-15
  • 0
    Actually you wrote 100 in the previous comment ... – 2017-01-15

2 Answers 2

1

$\displaystyle \sum^{50}_{r=0}r\cdot \binom{1000-r}{950}=0\cdot \binom{1000}{950}+1\cdot \binom{999}{950}+2\cdot \binom{998}{950}+\cdots \cdots \cdots +50\cdot \binom{950}{950}$

Using $\displaystyle \binom{n}{k} = $ Coefficients of $x^k$ in $(1+x)^n$

so coefficients of $\displaystyle x^{950}$ in

$\displaystyle 1\cdot (1+x)^{999}+2\cdot (1+x)^{998}+3\cdot (1+x)^{997}+\cdots \cdots +50 \cdot (1+x)^{950}$

let $S=1\cdot (1+x)^{999}+2\cdot (1+x)^{998}+3\cdot (1+x)^{997}+\cdots \cdots +50 \cdot (1+x)^{950}\cdots \cdots (\star)$

multiply both side by $\displaystyle \frac{1}{1+x}$

$\displaystyle S\cdot \frac{1}{1+x}=1\cdot (1+x)^{998}+2\cdot (1+x)^{997}+\cdots+49 \cdot (1+x)^{949}+50\cdot (1+x)^{949}\cdots \cdots (\star \star)$

So $\displaystyle \bigg(1-\frac{1}{1+x}\bigg) = (1+x)^{999}+(1+x)^{998}+\cdots \cdots (1+x)^{949}-50(1+x)^{949}$

So $\displaystyle S \cdot \frac{x}{1+x} = \frac{(1+x)^{1000}-(1+x)^{950}}{1+x-1}-50(1+x)^{949}$

So $\displaystyle S = \frac{(1+x)^{1001}-(1+x)^{951}}{x^2}-\frac{50(1+x)^{950}}{x}$

So coefficients of $x^{950}$ in $\displaystyle \frac{(1+x)^{1001}-(1+x)^{951}}{x^2}-$ coefficient of $x^{950}$ in $\displaystyle \frac{50(1+x)^{950}}{x}$

So coefficients of $x^{952}$ in $\bigg((1+x)^{1001}-(1+x)^{951}\bigg)-$ coefficients of $x^{951}$ in $(1+x)^{950}$

So we are getting $\displaystyle = \binom{1001}{952}$

1

Set $50-r=u$ $$\sum_{u=0}^{50}(51-u)\binom{950+u}{950}=\sum_{u=0}^{50}\{1002-(951+u)\}\binom{950+u}{950}$$

$$=1002\sum_{u=0}^{50}\binom{950+u}{950}-951\sum_{u=0}^{50}\binom{951+u}{951}$$

Now $\displaystyle\sum_{u=0}^{50}\binom{950+u}{950}$ is the coefficient of $x^{950}$ in $$\displaystyle\sum_{u=0}^{50}(1+x)^{950+u}$$