Solve the following questions:

Question

Following are two questions and their respective answers(as given in textbook):

Q1. If $$ k=\sum_{r=0}^n \frac{1}{n \choose r}$$, then write $$\sum_{r=0}^n \frac{r}{n \choose r}$$ in terms of k.

Ans. $$\frac {nk}{2}$$

Q2. If $$x+y=1$$ find $$\sum_{r=0}^n r^{2} y^{n-r} x^r$$.

Ans. $$n(nx+y)$$

The thing is, I have tried solving both the problems, but that hasn't been much fruitful. I faced difficulties from the first one as the expressions are sum of reciprocals of binomial coefficients. Because I never came across such reciprocals earlier. The problem with the second is because of the two terms x and y. Had either x or y not been there, I could have easily differentiated $$ (1+x)^n$$ or $$ (1+y)^n$$ and their respective expansions and derived the result.

So, I would be thankful to anyone who provides a proper solution for each of the above problems. Note: I wanted to add 'homework' tag, but wasn't allowed to do so.

Answer 1

For the first problem, $$k=\sum_{r=0}^n \frac{1}{n \choose r}$$ =$$\frac{1}{n \choose 0}+\frac{1}{n \choose 1}+..\frac{1}{n \choose n}$$ Now, $$ {n\choose r} = {n\choose n-r}$$ (I can now make pairs of two terms taking the first term from the beginning and last term as my first pair, I can add them directly as their denominators are the same) $$\implies k= \frac{2}{n \choose 0}+\frac{2}{n \choose 1}+\cdots$$ hence $$\frac{k}{2}=\frac{1}{n \choose 0}+\frac{1}{n \choose 1}+\cdots$$ (I'm not worried whether n is odd or even here, as you will see later that I have no interest in evaluating this sum)

Consider $$l=\sum_{r=0}^n \frac{r}{n \choose r}$$ $$\implies l=\frac{0}{n \choose 0}+\frac{1}{n \choose 1}+\frac{2}{n \choose 2}+\cdots +\frac{n-2}{n \choose n-2}+\frac{n-1}{n \choose n-1}+\frac{n}{n \choose n}$$ (Again I pair the terms the way I had done previously) $$\implies l=\frac{n}{n \choose 0}+\frac{n}{n \choose 1}+\cdots$$ $$\implies l=\frac{nk}{2}$$

Answer 2

For the second one, start with:

$\begin{align} \sum_{0 \le r \le n} x^r y^{n - r} &= y^n \sum_{0 \le r \le n} \left( \frac{x}{y} \right)^r \\ &= y^n \frac{1 - (x/y)^{n + 1}}{1 - x/y} \\ &= \frac{y^{n + 1} - x^{n + 1}}{y - x} \\ x \frac{\mathrm{d}}{\mathrm{d} x} \left( x \frac{\mathrm{d}}{\mathrm{d} x} \frac{y^{n + 1} - x^{n + 1}}{y - x} \right) &= \sum_{0 \le r \le n} r^2 x^r y^{n - r} \end{align}$

The result is a quite ugly expression:

$\begin{align} &\frac{y^n (x^2 y + x y^2) - (n + 1)^2 x^{n + 1} y^2 + (2 n^2 + 2 n - 1) x^{n + 2} y - n^2 x^{n + 3}} {(y - x)^3} \\ &\qquad = \frac{y^{n + 1} x (x + y) - x^{n + 1} ((n^2 + 2 n + 1) y^2 - (2 n^2 + 2 n - 1) x y + n^2 x^2)} {(y - x)^3} \\ &\qquad = \frac{y^{n + 1} x (x + y) - x^{n + 1} (y - x) ((n + 1)^2 y - n^2 x) } {(y - x)^3} \\ &\qquad = \frac{y^{n + 1} x - x^{n + 1} (y - x) ((n + 1)^2 y - n^2 x) } {(y - x)^3} \end{align}$

Yes, it is fine to consider $x$ and $y$ as separate variables, as mentioned. The relation $x + y = 1$ can be used to simplify the end result, however.