Evaluate and prove by induction: $\sum k{n\choose k},\sum \frac{1}{k(k+1)}$

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Solutions Collecting From Web of "Evaluate and prove by induction: $\sum k{n\choose k},\sum \frac{1}{k(k+1)}$"

1) Take $\displaystyle f(x)= \sum_{i=0}^n\binom{n}{i} x^i=(x+1)^n$. Now consider $f'(1)$.

2) Use that $\frac{1}{(k-1)\cdot k} = \frac{1}{k-1} -\frac{1}{k}$.

For the first one, you’re asked to find

$$\sum\limits_{k=0}^n k{n \choose k}$$

Since the binomial coefficients have the $n-k$ symmetry, we can put

$$\sum\limits_{k=0}^n (n-k){n \choose n-k}$$

thus

$$S_n = \sum\limits_{k=0}^n k{n \choose k}=\sum\limits_{k=0}^n (n-k){n \choose n-k}$$

But the RHS is

$$n\sum\limits_{k=0}^n {n \choose n-k}-\sum\limits_{k=0}^n k{n \choose n-k}$$

Now

$$S_n=n\sum\limits_{k=0}^n {n \choose k}-\sum\limits_{k=0}^n k{n \choose k}$$

or

$$S_n=n\sum\limits_{k=0}^n {n \choose k}-S_n$$

$$S_n=n2^n-S_n$$

$$2 S_n=n2^n $$

$$S_n=n2^{n-1} $$

The second one becomes easy once you make use of the telescoping property you’ve been suggested already.

The answer from lhf has already dealt with the second question posed here. Here is a Wikipedia article about it.

Here’s a probabilistic approach to the first question. When you toss a coin $n$ times, the probability that a “head” appears exactly $k$ times is $\dbinom n k (1/2)^n$. The average number of times a “head” appears is therefore
$$
\left( \binom n 0 \cdot 0 + \binom n 1 \cdot 1 + \binom n 2 \cdot 2 + \cdots + \binom n k \cdot k + \cdots + \binom n n \cdot n \right) \left(\frac 1 2 \right)^n.
$$
But the average number of times a “head” appears is obviously $n/2$. Therefore
$$
\left( \binom n 0 \cdot 0 + \binom n 1 \cdot 1 + \binom n 2 \cdot 2 + \cdots + \binom n k \cdot k + \cdots + \binom n n \cdot n \right) \left(\frac 1 2 \right)^n = \frac n 2.
$$
Consequently
$$
\binom n 0 \cdot 0 + \binom n 1 \cdot 1 + \binom n 2 \cdot 2 + \cdots + \binom n k \cdot k + \cdots + \binom n n \cdot n = n 2^{n-1} .
$$

If you don’t have to use induction, you can do this:

1)
$$
\sum_k k\binom{n}{k} = \sum_k k \frac{n(n-1)\cdots(n-k+1)}{k!} = n\sum_k \binom{n-1}{k-1} = n \sum_k \binom{n-1}{k} = n2^{n-1}
$$
Here $k$ runs over all integers and by convention $\binom{n}{k}$ is defined as zero if $k<0$ or $k > n$. The last equality follows from the fact that $\binom{n-1}{k}$ counts the $k$-element subsets of $\{1,\ldots,n-1\}$, so $\sum_k \binom{n-1}{k}$ counts the number of all subsets which is $2^{n-1}$.

2) Note that $\frac{1}{k(k+1)} = \frac{1}{k}-\frac{1}{k+1}$ so the sum is a telescope sum:
$$\sum_{k=1}^{n-1} \frac{1}{k(k+1)} = \sum_{k=1}^{n-1}\left( \frac{1}{k} – \frac{1}{k+1}\right) = 1 – \frac{1}{n}$$

Some different ways to prove $\sum_k k\binom{n}{k} = n2^{n-1}$ were suggested. I’ll add a combinatorial proof by double counting. Consider pairs $(a,A)$ were $A$ is a subset of $\{1,\ldots,n\}$ and $a \in A$. The number of such pairs is $\sum_k k\binom{n}{k}$ since there are $\binom{n}{k}$ ways to choose a $k$-element subset $A$ and then $k$ possibilities to choose $a \in A$. On the other hand, you can first choose $a \in \{1,\ldots,n\}$ and then add any subset of the remaining $n-1$ elements to make $A$, so this gives $n2^{n-1}$ possibilities. Comparing the two results shows that $\sum_k k\binom{n}{k} = n2^{n-1}$.

One way to do this:
1) You are looking for $\sum_{k=0}^n k\binom{n}{k}$.
First, look at $f(x)=(1+x)^n=\sum_{k=0}^n \binom{n}{k}x^k$, (by Newton’s binomial theorem). Hence $2^n=\sum_{k=0}^n \binom{n}{k}$, by calculating $f(1)$. Now, define $g(x)=\sum_{k=0}^n k\binom{n}{k}x^k$. What you need is $g(1)$. Try expressing $g(x)$ through $f(x)$. (Hint: what is $g'(x)$?)
2) Notice that $\frac{1}{(k-1)k}=\frac{1}{k-1}-\frac{1}{k}$. Hence:
$\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3}+\frac{1}{3\cdot 4} +\cdots+\frac{1}{(n-1)\cdot n}=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{n-1}-\frac{1}{n}\right)$

1)

$$S_n=\sum_{k=0}^nk\frac{n!}{k!(n-k)!}=n\sum_{k=1}^n\frac{(n-1)!}{(k-1)!(n-k)!}=n\sum_{k=0}^{n-1}\binom{n-1}k=n2^{n-1}.$$

Then the inductive proof by $S_0=0,S_n-S_{n-1}=T_n$, is a little tedious:
$$S_n-S_{n-1}
=\sum_{k=0}^nk\binom nk-\sum_{k=0}^{n-1}k\binom {n-1}k=n\binom nn+\sum_{k=0}^{n-1}k\left(\binom nk-\binom{n-1}k\right)\\
=n+\sum_{k=1}^{n-1}k\binom{n-1}{k-1}
=n+\sum_{k=1}^{n-1}(k-1)\binom{n-1}{k-1}+\sum_{k=1}^{n-1}\binom{n-1}{k-1}\\
=n+\sum_{k=1}^{n-1}(k-1)\binom{n-1}{k-1}+\sum_{k=1}^{n-1}\binom{n-1}{k-1}\\
=n+\sum_{k=0}^{n-2}k\binom{n-1}k+\sum_{k=0}^{n-2}\binom{n-1}{k}\\
=n+S_{n-1}-(n-1)+2^{n-1}-1=S_{n-1}+2^{n-1}.$$

And on the other hand

$$T_n=S_n-S_{n-1}=n2^{n-1}-(n-1)2^{n-2}=(n-1)2^{n-2}+2^{n-1}$$

2)

As $\dfrac1{k(k+1)}=\dfrac1k-\dfrac1{k+1}$, the sum telescopes as

$$S_n=\frac1{n+1}-1.$$

This time, the proof by induction is trivial:

$$S_0=0,$$

$$T_n=S_n-S_{n-1}=\frac1{n+1}-1-\frac1n+1=\frac1{n(n+1)}.$$