Given:
First term of a geometric series = a
Common ratio of the geometric series = [tex]\sqrt{3}[/tex]
To prove:
[tex]S_{10}=121a(\sqrt{3}+1)[/tex]
Solution:
The sum of first n terms of geometric series is
[tex]S_n=\dfrac{a(r^n-1)}{r-1}[/tex]
Where, a is the first term and r is the common ratio.
Putting n=10 and [tex]r=\sqrt{3}[/tex] in the above formula, to get the sum of 10 terms.
[tex]S_{10}=\dfrac{a((\sqrt{3})^{10}-1)}{\sqrt{3}-1}[/tex]
[tex]S_{10}=\dfrac{a(243-1)}{\sqrt{3}-1}[/tex]
[tex]S_{10}=\dfrac{242a}{\sqrt{3}-1}[/tex]
Rationalizing the denominator, we get
[tex]S_{10}=\dfrac{242a}{\sqrt{3}-1}\times \dfrac{\sqrt{3}+1}{\sqrt{3}+1}[/tex]
[tex]S_{10}=\dfrac{242a(\sqrt{3}+1)}{(\sqrt{3})^2-1^2}[/tex]
[tex]S_{10}=\dfrac{242a(\sqrt{3}+1)}{3-1}[/tex]
[tex]S_{10}=\dfrac{242a(\sqrt{3}+1)}{2}[/tex]
[tex]S_{10}=121a(\sqrt{3}+1)[/tex]
Hence proved.
