Notice that
13 - 9 = 4
17 - 13 = 4
so it's likely that each pair of consecutive terms in the sum differ by 4. This means the last term, 149, is equal to 9 plus some multiple of 4 :
149 = 9 + 4k
140 = 4k
k = 140/4
k = 35
This tells you there are 35 + 1 = 36 terms in the sum (since the first term is 9 plus 0 times 4, and the last term is 9 plus 35 times 4). Among the given options, only the first choice contains the same amount of terms.
Put another way, we have
[tex]\displaystyle 9 + 13 + 17 + \cdots + 149 = \sum_{k=0}^{35} (9 + 4k)[/tex]
but if we make the sum start at k = 1, we need to replace every instance of k with k - 1, and accordingly adjust the upper limit in the sum.
[tex]\displaystyle 9 + 13 + 17 + \cdots + 149 = \sum_{k-1=0}^{35+1} (9 + 4(k-1))[/tex]
[tex]\displaystyle 9 + 13 + 17 + \cdots + 149 = \sum_{k=1}^{36} (5 + 4k)[/tex]
