The sum of $n$ terms of the series $12 + 16 + 24 + 40 + \dots$ is

  • A
    $2(2^n - 1) + 8n$
  • B
    $2(2^n - 1) + 6n$
  • C
    $3(2^n - 1) + 8n$
  • D
    $4(2^n - 1) + 8n$

Explore More

Similar Questions

$\frac{1^{2}}{2} + \frac{1^{2}+2^{2}}{3} + \frac{1^{2}+2^{2}+3^{2}}{4} + \frac{1^{2}+2^{2}+3^{2}+4^{2}}{5} + \dots$ up to $8$ terms $=$

$1 + \frac{1^3 + 2^3}{1 + 2} + \frac{1^3 + 2^3 + 3^3}{1 + 2 + 3} + \dots + \frac{1^3 + 2^3 + 3^3 + \dots + 15^3}{1 + 2 + 3 + \dots + 15} - \frac{1}{2}(1 + 2 + 3 + \dots + 15)$ is equal to

If $S_{n} = 4 + 11 + 21 + 34 + 50 + \ldots$ to $n$ terms,then $\frac{1}{60}(S_{29} - S_{9})$ is equal to $.......$.

If the set of natural numbers is partitioned into subsets $S_1 = \{1\}, S_2 = \{2, 3\}, S_3 = \{4, 5, 6\}$ and so on,then the sum of the terms in $S_{50}$ is

Difficult
View Solution

The sum of the series $1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + \dots$ to $n$ terms is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo