Question

${ }^{n}C_1 - (1+\frac{1}{2}){ }^{n}C_2 + (1+\frac{1}{2}+\frac{1}{3}){ }^{n}C_3 - (1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}){ }^{n}C_4 + ... + (-1)^{n-1}(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}){ }^{n}C_n =$

• A. $\frac{n-1}{n}$
• B. $\frac{1}{n}$
• C. $\frac{1}{n+1}$
• D. $\frac{2^n}{n+1}$

Answer 1

Answer: (B)

$\frac{1}{n}$

Answer 2

We are given

$$S=\sum_{k=1}^{n}(-1)^{k-1}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{k}\right)\binom{n}{k}.$$

Notice that the harmonic number $H_k=\sum_{j=1}^{k}\frac{1}{j}$. Thus,

$$S=\sum_{k=1}^{n}(-1)^{k-1}H_k\binom{n}{k}=\sum_{k=1}^{n}(-1)^{k-1}\binom{n}{k}\left(\sum_{j=1}^{k}\frac{1}{j}\right).$$

Interchange the order of summation:

$$S=\sum_{j=1}^{n}\frac{1}{j}\sum_{k=j}^{n}(-1)^{k-1}\binom{n}{k}.$$

Using the known binomial identity

$$\sum_{k=j}^{n}(-1)^{k-1}\binom{n}{k}=(-1)^{j-1}\binom{n-1}{j-1},$$

we have

$$S=\sum_{j=1}^{n}\frac{(-1)^{j-1}}{j}\binom{n-1}{j-1}.$$

Let $m=j-1$ so that $m=0$ to $n-1$. Then,

$$S=\sum_{m=0}^{n-1}\frac{(-1)^m}{m+1}\binom{n-1}{m}.$$

It is known that

$$\sum_{m=0}^{N}\frac{(-1)^m}{m+1}\binom{N}{m}=\frac{1}{N+1},$$

where here $N=n-1$. Therefore,

$$S=\frac{1}{n}.$$