First slide

Special Series in Sequences and Series

Question

The sum of the series $1 \cdot 3 \cdot 5 + 2 \cdot 5 \cdot 8 + 3 \cdot 7 \cdot 11 + \dots$ up to n terms is

Moderate

A

$\frac{n (n + 1) (9 n^{2} + 23 n + 13)}{6}$
B

$\frac{n (n - 1) (9 n^{2} + 23 n + 12)}{6}$
C

$\frac{(n + 1) (9 n^{2} + 23 n + 13)}{6}$
D

$\frac{n (9 n^{2} + 23 n + 13)}{6}$

Solution

Let, $S_{n} = 1 \cdot 3 \cdot 5 + 2 \cdot 5 \cdot 8 + 3 \cdot 7 \cdot 11 + \dots + n th term$
$\begin{array}{l} ∴ T_{n} = n (2 n + 1) (3 n + 2) \\ ∴ S_{n} = {ΣT}_{n} = Σn (2 n + 1) (3 n + 2) \\ = Σn (6 n^{2} + 7 n + 2) \\ = Σ (6 n^{3} + 7 n^{2} + 2 n) \\ = 6 {Σn}^{3} + 7 {Σn}^{2} + 2 Σn \\ = 6 {[\frac{n (n + 1)}{2}]}^{2} + \frac{7 n (n + 1) (2 n + 1)}{6} + 2 \frac{n (n + 1)}{2} \\ = \frac{n (n + 1)}{2} [\frac{6 n (n + 1)}{2} + \frac{7 (2 n + 1)}{3} + 2] \\ = \frac{n (n + 1)}{2} [\frac{18 (n^{2} + n) + 28 n + 14 + 12}{6}] \end{array}$
$\begin{array}{l} = \frac{n (n + 1)}{2} (\frac{18 n^{2} + 46 n + 26}{6}) \\ = \frac{n (n + 1)}{2} \times \frac{2 (9 n^{2} + 23 n + 13)}{6} \\ = \frac{n (n + 1) (9 n^{2} + 23 n + 13)}{6} \end{array}$

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