Hello!
We have the following expression:
[tex]\sum_{n\mathop{=}0}^6300(1.05)^{n-1}[/tex]
We can solve it as a geometric sequence, I'll show you how.
First, let's write the number of terms of the sequence:
Starting in 0 until 6, we have: a1, a2, a3, a4, a5, a6 and a7.
(7 terms).
Also, we can see that the ratio is 1.05.
Now, let's calculate the first term of the sequence (a1):
[tex]\begin{gathered} a_1=300\times(1.05)^{0-1} \\ a_1=300\times(1.05)^{-1} \\ a_1=300\times\frac{1}{1.05} \\ a_1=\frac{2000}{7} \end{gathered}[/tex]
As we know the first term of the sequence and the ratio, we can use the formula below to calculate the sum of the 7 terms of this sequence:
[tex]\boxed{S_n=\frac{a_{1}(r^{n}-1)}{r-1}}[/tex]
So, let's replace it with the values that we already know:
[tex]S_7=\frac{\frac{2000}{7}(1.05^7-1)}{1.05-1}=\frac{\frac{2000}{7}(1.4071-1)}{0.05}=\frac{\frac{2000}{7}(0.4071)}{0.05}\cong2326.29[/tex]
Right answer: alternative B.
