Name: Finding Sum of Series Using Method of Differences
Uploaded: 2019-07-19
Duration: 7 min 14 s
Description: In this video, we will learn to solve problems based on the summation of series by using the method of differences. Grab a pen and paper, and try to solve th...

Question

Prove that sum of cubes of any number of consecutive natural numbers is always divisible by the sum of these numbers.

Accepted Answer

Let the $m$ consecutive natural numbers are $n + 1, n + 2, n + 3, \dots, n + m$ .

$∴ S_{n} = (n + 1)^{3} + (n + 2)^{3} \dots \dots + (n + m)^{3} = \sum_{k = 1}^{m} (n + k)^{3}$

From the properties of sigma notation, we get

$S_{n} = \sum_{i = 1}^{m} n^{3} + \sum_{k = 1}^{m} k^{3} + 3 n^{2} \sum_{k = 1}^{m} k + 3 n \sum_{i = 1}^{n} k^{2}$

We know that, sum of cubes of first $n$ terms is $\frac{n^{2} (n + 1)^{2}}{4}$ .

Sum of the squares of the first $n$ terms is $\frac{n (n + 1) (2 n + 1)}{6}$ .

Sum of the first $n$ terms is $\frac{n (n + 1)}{2}$ .

$\Rightarrow S_{n} = n^{3} m + \frac{m^{2} (m + 1)^{2}}{4} + \frac{3 n^{2} m (m + 1)}{2} + \frac{3 n m (m + 1) (2 m + 1)}{6}$

$\Rightarrow S_{n} = \frac{m}{4} [4 n^{3} + m (m + 1)^{2} + 6 n^{2} (m + 1) + 2 n (m + 1) (2 m + 1)]$

$= \frac{m}{4} [4 n^{3} + 4 n^{2} (m + 1) + 2 n^{2} (m + 1) + 2 n (m + 1) (2 m + 1) + m (m + 1)^{2}]$

$= \frac{m}{4} [4 n^{2} (n + m + 1) + 2 n (m + 1) (n + m + 1) (3 n m (m + 1) + m (m + 1)^{2}]$

$= \frac{m}{4} [(n + m + 1) 2 n (2 n + m + 1) + η (m - \cup (2 n + m + 1)]$

$= \frac{m}{4} (2 n + m + 1) [4 n^{2} + 2 n (m + 4 + m (m + 1)]$ (i)

Also, $S_{n}^{'} = (n + 1) + (n + 2) \dots + n$

$= \sum_{k = 1}^{n} (n + k) = n \cdot m + \frac{m (m + 1)}{2} = \frac{m}{2} (2 n + m + 1)$ (ii)

From (i) and (ii) we get

Thus, sum of cubes of any number of consecutive natural numbers is always divisible by the sum of these numbers.

Hence, proved.

Prove that sum of cubes of any number of consecutive natural numbers is always divisible by the sum of these numbers.

Important Questions on Sequences and Series

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