First slide

Series of natural numbers

Question

Sum to $n$ terms of the series $\frac{1}{1 \cdot 2 \cdot 3 \cdot 4} + \frac{1}{2 \cdot 3 \cdot 4 \cdot 5} + \frac{1}{3 \cdot 4 \cdot 5 \cdot 6} + \dots$ is

Moderate

A

$\frac{1}{24} - \frac{1}{n}$
B

$\frac{1}{18} - \frac{1}{3 (n + 1) (n + 2) (n + 3)}$
C

$\frac{1}{2^{n}} - \frac{1}{n^{2}}$
D

$\frac{1}{18} - \frac{1}{3 (n + 3) (n + 4)}$

Solution

Let $t_{r}$ denote the $r t h$ term of the given series, then
   $t_{r} = \frac{1}{r (r + 1) (r + 2) (r + 3)}$ and
   $t_{r + 1} = \frac{1}{(r + 1) (r + 2) (r + 3) (r + 4)}$
$\Rightarrow r t_{r} = \frac{1}{(r + 1) (r + 2) (r + 3)}$ and

   $(r + 4) t_{r + 1} = \frac{1}{(r + 1) (r + 2) (r + 3)}$
Thus, $r t_{r} - (r + 4) t_{r + 1} = 0$
$\Rightarrow r t_{r} - (r + 1) t_{r + 1} = 3 t_{r + 1}$
Putting $r = 1, 2, \dots, n - 2, n - 1$ , we get
$\begin{array}{l} 1 t_{1} - 2 t_{2} = 3 t_{2} \\ 2 t_{2} - 3 t_{3} = 3 t_{3} \\ 3 t_{3} - 4 t_{4} = 3 t_{4} \\ \dots \\ (n - 2) t_{n - 2} - (n - 1) t_{n - 1} = 3 t_{n - 1} \\ (n - 1) t_{n - 1} - n t_{n} = 3 t_{n} \end{array}$
Adding above expressions, we get
$\begin{array}{l} t_{1} - n t_{n} = 3 (t_{2} + t_{3} + \dots + t_{n}) \\ \Rightarrow 4 t_{1} - n t_{n} = 3 (t_{1} + t_{2} + \dots + t_{n}) \\ \Rightarrow \frac{1}{1 \cdot 2 \cdot 3} - \frac{1}{(n + 1) (n + 2) (n + 3)} = 3 S_{n} \\ \Rightarrow S_{n} = \frac{1}{3} [\frac{1}{6} - \frac{1}{(n + 1) (n + 2) (n + 3)}] \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App