# How do I find the sum of the series: 4+5+6+8+9+10+12+13+14+⋯+168+169+170. since D is changing from +1, +1 to +2 ?

Oct 20, 2015

$10962$, see the explanation.

#### Explanation:

If we make three sums:

$4 + 8 + 12 + \ldots = {\sum}_{k = 1}^{n} 4 k$
$5 + 9 + 13 + \ldots = {\sum}_{k = 1}^{n} \left(4 k + 1\right) = {\sum}_{k = 1}^{n} 4 k + {\sum}_{k = 1}^{n} 1 = {\sum}_{k = 1}^{n} 4 k + n$
$6 + 10 + 14 + \ldots = {\sum}_{k = 1}^{n} \left(4 k + 2\right) = {\sum}_{k = 1}^{n} 4 k + {\sum}_{k = 1}^{n} 2 = {\sum}_{k = 1}^{n} 4 k + 2 n$

Then :

$\sum = {\sum}_{k = 1}^{n} 4 k + {\sum}_{k = 1}^{n} 4 k + n + {\sum}_{k = 1}^{n} 4 k + 2 n = 3 {\sum}_{k = 1}^{n} 4 k + 3 n$

$\sum = 12 {\sum}_{k = 1}^{n} k + 3 n = 12 \frac{n \left(n + 1\right)}{2} + 3 n = 6 n \left(n + 1\right) + 3 n$

$\sum = 3 n \left(2 \left(n + 1\right) + 1\right) = 3 n \left(2 n + 3\right)$

Note: sum of first $n$ integers is $\frac{n \left(n + 1\right)}{2}$.

We have to find $n$ and it's the number of members in the first sum, so:

$n = \frac{168}{4} = 42$

Finally:

$\sum = 3 \cdot 42 \cdot \left(2 \cdot 42 + 3\right) = 126 \cdot 87 = 10962$