Question

The sum of n terms of the series is 1/√3+1+3/√3+......
options:-
(a) (1/6) (3+√3) (3^(n/2) - 1)
(b) (1/6) (√3+1) (3^(n/2) - 1)
(c) (1/6) (3+√3) (3^(n/2) + 1)
(d) none​

Answer 1

Answer:

Answer: The required sum is \dfrac{9841(3+\sqrt3)}{3}.

3

9841(3+

3

)

.

Step-by-step explanation: We are given to find the sum of first 18 terms of the following series :

$$\dfrac{1}{\sqrt3},~1,~\dfrac{3}{\sqrt},~.~~.~~.$$

If a(n) denotes the nth term of the given series, then we see that

$$a(n)=\dfrac{1}{\sqrt3},~~a(2)=1,~~a(3)=\dfrac{3}{\sqrt3},~~.~~.~~.$$

That is,

$$\dfrac{a(2)}{a(1)}=\dfrac{a(3)}{a(2)}=\dfrac{1}{\frac{1}{\sqrt3}}=\dfrac{\frac{3}{\sqrt3}}{1}=\sqrt3.$$

So, the given series is a geometric series with first term and common ratio as follows :

$$a=\dfrac{1}{\sqrt3},~~r=\sqrt3.$$

We know that

the sum of first n terms with first term a and common ratio r, with |r| > 1 is given by

$$S_r=\dfrac{a(r^n-1)}{r-1}.$$

Therefore, the sum of first 18 terms of the given series is

$$\begin{gathered}S_{18}\\\\\\=\dfrac{\frac{1}{\sqrt3}((\sqrt3)^{18}-1)}{\sqrt3-1}\\\\\\=\dfrac{1}{\sqrt3(\sqrt3-1)}(3^9-1)\\\\\\=\dfrac{1}{\sqrt3(\sqrt3-1)}(19683-1)\\\\\\=\dfrac{19682}{\sqrt3(\sqrt3-1)}\\\\\\=\dfrac{19682(\sqrt3+1)}{\sqrt3(3-1)}\\\\\\=\dfrac{9841(\sqrt3+1)}{\sqrt3}\\\\\\=\dfrac{9841(3+\sqrt3)}{3}.\end{gathered}$$

Thus, the required sum is $$\dfrac{9841(3+\sqrt3)}{3}.$$