Name: Problems on General Term of an A.P
Uploaded: 2019-07-19
Duration: 5 min 14 s
Description: In this video, we are going to learn how to find the nth term of an AP and an example problem to learn how to find the nth term of an arithmetic progression.

Question

Find the sum of the series $(2^{3} - 1) + (4^{3} - 3^{3}) + (6^{3} - 15^{3}) + \dots$ to $n$ terms.

Accepted Answer

Given series $(2^{3} - 1) + (4^{3} - 3^{3}) + (6^{3} - 15^{3}) + \dots \dots \dots n$

$= 2^{3} + 4^{3} + 6^{3} + \dots . . n terms - (1^{3} + 3^{3} + 5^{3} + \dots n terms)$

$= \sum_{1}^{n} [(2 n)^{3} - (2 n - 1)^{3}]$

We know $a^{3} - b^{3} = (a - b)^{3} + 3 a b (a - b)$

$= \sum_{1}^{n} [(2 n - 2 n + 1)^{3} + 3 (2 n) (2 n - 1) (2 n - 2 n + 1)]$

$= \sum_{1}^{n} [1^{3} + 3 (2 n) (2 n - 1)]$

$= \sum_{1}^{n} [1 + 6 n (2 n - 1)]$

$= \sum_{1}^{n} [1 + 12 n^{2} - 6 n]$

$= \sum_{1}^{n} (1 + 12 n^{2} - 6 n)$

$= \sum_{1}^{n} (1) + \sum_{1}^{n} (12 n^{2}) + \sum_{1}^{n} (- 6 n)$

$= n + 12 \cdot \frac{n (n + 1) (2 n + 1)}{6} - 6 \cdot \frac{n (n + 1)}{2}$

$= n + 2 n (n + 1) (2 n + 1) - 3 n (n + 1)$

$= n + (2 n^{2} + 2 n) (2 n + 1) - 3 n^{2} - 3 n$

$= n + 4 n^{3} + 4 n^{2} + 2 n^{2} + 2 n - 3 n^{2} - 3 n$

$= 4 n^{3} + 3 n^{2}$

Hence, the required sum is $4 n^{3} + 3 n^{2}$ .

Find the sum of the series $(2^{3} - 1) + (4^{3} - 3^{3}) + (6^{3} - 15^{3}) + \dots$ to $n$ terms.

Important Questions on Numbers and Sequences

Learn and Prepare for any exam you want !

Find the sum of the series 23-1+43-33+63-153+… to n terms.

Important Questions on Numbers and Sequences

Learn and Prepare for any exam you want !

Find the sum of the series $(2^{3} - 1) + (4^{3} - 3^{3}) + (6^{3} - 15^{3}) + \dots$ to $n$ terms.