Question

A sequence $d_1,d_2,d_3,\cdots$ is defined by letting $d_1=2$ and $d_k=\frac{d_{k-1}}{k}$ for all natural number $k\ge 2.$ Then which of the following is/are correct?
(Solve using mathematical induction)

• A. $d_n=\frac{2}{n(n-1)}$
• B. $d_n=\frac{2}{n!}$
• C. $d_3-4d_4=0$
• D. $\frac{1}{d_5}-(\frac{1}{d_3}+\frac{1}{d_4})=1$

Answer 1

Answer: (B), (C)

$d_3-4d_4=0$

From mathematical induction : sequence is true for $k=2,3,4,\cdots$
For $k=2,d_2=\frac{d_1}{2}$
For $k=3,d_3=\frac{d_2}{3}=\frac{d_1}{2\cdot 3}$
For $k=4,d_4=\frac{d_3}{4}=\frac{d_1}{2\cdot 3\cdot 4}$
Similarly, let $d_n=\frac{d_1}{1\cdot 2\cdot 3\cdots n}$ is true for $k=n$
now check at $n=k+1,d_{n+1}=\frac{d_n}{n+1}$
$\Rightarrow d_{n+1}=\frac{d_1}{1\cdot 2\cdot 3\cdots n}\cdot \frac{1}{n+1}\Rightarrow d_{n+1}=\frac{d_1}{1\cdot 2\cdot 3\cdots n\cdot (n+1)}$ is true for $k=n+1$
So, $d_n=\frac{d_1}{1\cdot 2\cdot 3\cdots n}=\frac{2}{n!}$ is true