Find the sum of the series

Solution:

Given $\left(3^{3}-2^{3}\right)+\left(5^{3}-4^{3}\right)+\left(7^{3}-6^{3}\right)+\ldots$

Let the series be $S=\left(3^{3}-2^{3}\right)+\left(5^{3}-4^{3}\right)+\left(7^{3}-6^{3}\right)+\ldots$

i) Generalizing the series in terms of i

$\mathrm{S}=\sum_{\mathrm{i}=1}^{\mathrm{n}}\left[(2 \mathrm{i}+1)^{3}-(2 \mathrm{i})^{3}\right]$

Using the formula $a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)$

$\Rightarrow S=\sum_{i=1}^{n}(2 i+1-2 i)\left((2 i+1)^{2}+(2 i+1)(2 i)+(2 i)^{2}\right)$

$\Rightarrow S=\sum_{i=1}^{n}\left(4 i^{2}+4 i+1+4 i^{2}+2 i+4 i^{2}\right)$

On simplifying and computing we get

$\Rightarrow S=\sum_{i=1}^{n}\left(12 i^{2}+6 i+1\right)$

Now by splitting the summation we get

$\Rightarrow S=12 \sum_{i=1}^{n} i^{2}+6 \sum_{i=1}^{n} i+\sum_{i=1}^{n} 1$

We know that $\sum n^{2}=\frac{n(n+1)(2 n+1)}{6}$ and $\sum n=\frac{n(n+1)}{2}$

$\Rightarrow S=\sum_{i=1}^{n}\left(12 i^{2}+6 i+1\right)$

Now by splitting the summation we get

$\Rightarrow S=12 \sum_{i=1}^{n} i^{2}+6 \sum_{i=1}^{n} i+\sum_{i=1}^{n} 1$

We know that $\sum n^{2}=\frac{n(n+1)(2 n+1)}{6}$ and $\sum n=\frac{n(n+1)}{2}$

Using the above formula we get

$\Rightarrow \mathrm{S}=12 \frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}+6 \frac{\mathrm{n}(\mathrm{n}+1)}{2}+\mathrm{n}$

Simplifying we get

⇒ S = 2n (n+1) (2n+1) + 3n (n+1) + n

⇒ S = 2n (2n2 + 2n + n + 1) + 3n2 + 3n + n

⇒ S = 4n3 + 6n2 + 2n + 3n2 + 4n

⇒ S = 4n3 + 9n2 + 6n

Hence sum up to n terms is 4n3 + 9n2 + 6n

ii) Sum of first 10 terms or up to 10 terms

To find sum up to 10 terms put n = 10 in S

⇒ S = 4 (10)3 + 9 (10)2 + 6 (10)

⇒ S = 4000 + 900 + 60

⇒ S = 4960

Hence sum of series up to 10 terms is 4960