# Question b2691

Sep 20, 2016

$\text{Sum=} 736$.

#### Explanation:

Let $S$ be the sum. Then, rearranging the terms of the series,

$S = \left[1 + 4 + 7 + 10 + 13 + \ldots\right] + \left[3 + 6 + 9 + 12 + \ldots + 45\right]$

=U+V, say, where,

U=[1+4+7+10+13+... ], &,

$V = \left[3 + 6 + 9 + 12 + \ldots + 45\right]$

We notice that the General ${n}^{t h}$ Term ${u}_{n}$ of the Series $U$ is,

${u}_{n} = 1 + \left(n - 1\right) 3 = 3 n - 2 \text{, whereas, that for V is } {v}_{n} = 3 n$.

Clearly, ${u}_{16} = 46 , \mathmr{and} , {v}_{15} = 45$.

Therefore, $U = {\sum}_{n = 1}^{16} {u}_{n} = {\sum}_{n = 1}^{16} \left(3 n - 2\right)$

$= 3 {\sum}_{n = 1}^{16} n - 2 {\sum}_{n = 1}^{16} 1$

$= 3 {\left[\frac{1}{2} n \left(n + 1\right)\right]}_{n = 1}^{16} - 2 \left(16\right) = \frac{3}{2} \cdot 16 \cdot 17 - 32 = 408 - 32 = 376$

Similarly, $V = 360$.

Finally, therefore, $S = 376 + 360 = 736.$

In fact, $U$ can easily obtained by using the formula

$U = {\sum}_{n = 1}^{16} {u}_{n} = \left[\frac{n}{2} \cdot \text{(first term+last term)}\right]$

$= \frac{16}{2} \left(1 + 46\right) = 8 \cdot 47 = 376 , \text{as before!}$