10. The sum of the series
= 1/square root of 3 + 1 + 3/ square root of 3 ......
.....to 18 terms is
Answers
Step-by-step explanation:
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},~~.~~.~~.a(n)=
3
1
, a(2)=1, a(3)=
3
3
, . . .
That is,
\dfrac{a(2)}{a(1)}=\dfrac{a(3)}{a(2)}=\dfrac{1}{\frac{1}{\sqrt3}}=\dfrac{\frac{3}{\sqrt3}}{1}=\sqrt3.
a(1)
a(2)
=
a(2)
a(3)
=
3
1
1
=
1
3
3
=
3
.
So, the given series is a geometric series with first term and common ratio as follows :
a=\dfrac{1}{\sqrt3},~~r=\sqrt3.a=
3
1
, r=
3
.
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}.S
r
=
r−1
a(r
n
−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}
S
18
=
3
−1
3
1
((
3
)
18
−1)
=
3
(
3
−1)
1
(3
9
−1)
=
3
(
3
−1)
1
(19683−1)
=
3
(
3
−1)
19682
=
3
(3−1)
19682(
3
+1)
=
3
9841(
3
+1)
=
3
9841(3+
3
)
.
Thus, the required sum is \dfrac{9841(3+\sqrt3)}{3}.
3
9841(3+
3
)
.