Find the sum of series.

Question

1000C50 +2.999C49 +3.998C48 +......... +51.950C0

$$\binom{1000}{50}+2\cdot \binom{999}{49}+3\cdot \binom{998}{48}+\cdots\cdots \cdots \cdots +51\cdot \binom{950}{0} $$

I tried to apply the pascal rule but failed as there are some numbers with the binomial coefficients, any hint would be helpful thanks.

Answer 1

A purely combinatorial solution is possible, though it takes some experience to see it before knowing what the sum is. Let $S=\{0,1,\ldots,1001\}$. You want to count the subsets of $S$ having $952$ elements.

Suppose that $A$ is such a subset, and $k$ is its second-smallest element: there is exactly one member of $A$ smaller than $k$, and $950$ members of $A$ are larger than $k$. There are $k$ elements of $S$ smaller than $k$, so there are $k$ ways to choose the one member of $A$ that is smaller than $k$. There are $1001-k$ elements of $S$ bigger than $k$, so there are $\binom{1001-k}{950}$ ways to choose the $950$ elements of $A$ that are bigger than $k$. Thus, there are

$$k\binom{1001-k}{950}$$

$950$-element subsets of $S$ whose second-largest member is $k$. The smallest possible value of $k$ is $1$, since there has to be at least one member of $S$ smaller than $k$, and the largest is $51$, since there have to be at least $950$ members of $S$ larger than $k$, so the total number of $950$-element subsets of $S$ is

$$\begin{align*} \sum_{k=1}^{51}k\binom{1001-k}{950}&=\sum_{k=1}^{50}k\binom{1001-k}{(1001-k)-950}\\ &=\sum_{k=1}^{51}k\binom{1001-k}{51-k}\\ &=\binom{1000}{50}+2\binom{999}{49}+3\binom{998}{48}+\ldots+50\binom{951}1+51\binom{950}0\;. \end{align*}$$

On the other hand, there are $\binom{1002}{950}=\binom{1002}{52}$ subsets of $S$ with $950$ elements, so

$$\binom{1000}{50}+2\binom{999}{49}+3\binom{998}{48}+\ldots+50\binom{951}1+51\binom{950}0=\binom{1002}{952}\;.$$

Answer 2

Using $\displaystyle \binom{n}{r} = \binom{n}{n-r}$, we can write it as

$$\binom{1000}{950}+2\cdot \binom{999}{950}+3\cdot \binom{998}{950}+\cdots+51\cdot \binom{950}{950}\;,$$

which is same as finding the coefficient of $x^{50}$ in

$$(1+x)^{1000}+2(1+x)^{999}+3(1+x)^{998}+\cdots+51(1+x)^{950}\;.$$

Now let

$$S = (1+x)^{1000}+2(1+x)^{999}+3(1+x)^{998}+\cdots+51(1+x)^{950}\tag{1}\;,$$

so that

$$\frac{S}{1+x} = (1+x)^{999}+2(1+x)^{998}+3(1+x)^{997}+\cdots+51\cdot (1+x)^{949}\tag{2}\;.$$

Then subtract $(2)$ from $(1)$:

$$\begin{align*} S-\frac{S}{1+x}&=\left(1-\frac1{1+x}\right)S=\frac{x}{1+x}S\\\\ &= \underbrace{(1+x)^{1000}+(1+x)^{999}+(1+x)^{998}+\cdots+(1+x)}_{\bf{Geometric\; Progression \; with \; common \; ratio\; (1+x) }}-51(1+x)^{949}\;, \end{align*}$$

so

$$\frac{x}{1+x}\cdot S = (1+x)\cdot \frac{(1+x)^{1000}-1}{x}-51(1+x)^{949}\;,$$

and

$$S = \frac{(1+x)^{1002}}{x^2}-\frac{(1+x)^2}{x^2}-\frac{51(1+x)^{950}}{x^2}\;.$$

Thus, we want the coefficient of $x^{950}$ in

$$\frac{(1+x)^{1002}}{x^2}-\frac{(1+x)^2}{x^2}-\frac{51(1+x)^{950}}{x^2}\;,$$

which is the coefficient of $x^{952}$ in $(1+x)^{1002}-(1+x)^2-51(1+x)^{950}$, and that is $\dbinom{1002}{952}$.

Answer 3

A shorter solution! $$\begin{align} \sum_{r=0}^{50}(r+1)\binom {1000-r}{50-r} &=\sum_{r=0}^{50}(r+1)\binom {1000-r}{950}\\ &=\sum_{s=950}^{1000}(1001-s)\binom s{950}&&(s=1000-r)\\ &=\sum_{s=950}^{1000}\sum_{t=s}^{1000}\binom s{950}\\ &=\sum_{t=950}^{1000}\sum_{s=950}^t\binom s{950} &&(950\le s\le t\le 1000)\\ &=\sum_{t=950}^{1000}\binom {t+1}{951}\\ &=\color{red}{\binom {1002}{952}=\binom {1002}{50}} \end{align}$$

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Previous answer below. $$\scriptsize\begin{align} \sum_{r=0}^{50}(r+1)\binom {1000-r}{50-r} &=\sum_{r=0}^{50}(r+1)\binom {1000-r}{950}\\ &=\sum_{s=950}^{1000}(1001-s)\binom s{950}&&(s=1000-r)\\ &=1002\sum_{s=950}^{1000}\binom s{950}-\sum_{s=950}^{1000}(s+1)\binom s{950}\\ &=1002\binom {1001}{951}-951\sum_{s=950}^{1000}\binom {s+1}{951}\\ &=952\binom {1002}{952}-951\binom {1002}{952}\\ &=\color{red}{\binom{1002}{952}=\binom {1002}{50}}\end{align}$$