The correct answer is option B which is Sn = -111974.
What is geometric expansion?
When there is a constant between the two successive numbers in the series then it is called a geometric series.
Here we have the following data:-
[tex]\sum_{n=1}^7-2.6^{n-1}[/tex]
[tex]a_n = a_1r^{n-1}\ \ \ \ \ S_n = a_1(\dfrac{1-r^n}{1-r})[/tex]
Here we need to calculate the first term second term and common ratio to calculate the sum of the series.
a₁ = -2.(6¹⁻¹) = -2
a₂ = -2.( 6²⁻¹) = -2 (6) =-12
Common ratio:-
[tex]a_n = a_1r^{n-1}[/tex]
[tex]-12 = -2r^{2-1}[/tex]
r = 6
Put all the values in the sum formula:-
[tex]S_n = -2(\dfrac{1-6^7}{1-6})[/tex]
[tex]S_n = -2(\dfrac{279935}{5})=-2\times 55987[/tex]
[tex]S_n[/tex] = -111974
Therefore the correct answer is option B which is Sn = -111974.
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