Question

2+6+18+....+2.3n-1=(3n-1)

Answer 1

Explanation:

Given that

2+6+18+....+2.3n-1=(3n-1)

We have to prove it.

2+6+18+....+2.3n-1   is a Geometric series.

Here

a₁ = 2

r = 6/2 = 3

r = 3           so    |r| > 1

We have to show that Sum of "n" terms is equal to "3^n-1"

We know that

$S_{n}=\frac{a_{1}(r^{n}-1)}{r-1}\\\\S_{n}=\frac{2(3^{n}-1)}{3-1}\\ \\S_{n}=\frac{2(3^{n}-1)}{2}\\\\S_{n}=3^{n}-1$

Which is the required answer.

I hope it will help you.

Answer 2

Answer:

Step-by-step explanation:

$2+6+18..............+2{(3)}^{n-1}$

since each term of this series is obtained by multiplying 3 to the immediately preceeding term, this is a geometric series with common ratio 3.

Here, a=2, r=3

$2+6+18..............+2{(3)}^{n-1}$

$=\frac{a(r^{n}-1)}{r-1}$

$=\frac{2(3^{n}-1)}{3-1}$

$=\frac{2(3^{n}-1)}{2}$

$=3^{n}-1$