Answer:
Option (d) is correct.
6 + 9 + 12 +15 = [tex]\sum_{k=1}^{4}(3k+3)[/tex]
Step-by-step explanation:
Consider the given series expansion 6 + 9 + 12 +15.
since we have to write in series form of 3.
We first write each term of given series in term of 3k or 3k+3
6 can be written as 3(2) or 3(1)+3
9 can be written as 3(3) or 3(2)+3
12 can be written as 3(4) or 3(3)+3
15 can be written as 3(5) or 3(4)+3
Thus, we can write the given series as ,
[tex]\sum_{k=2}^{5}3k[/tex] or [tex]\sum_{k=1}^{4}(3k+3)[/tex]
Since, we have options to choose.
From the given option (d) matches our result.
So, 6 + 9 + 12 +15 = [tex]\sum_{k=1}^{4}(3k+3)[/tex]
