The sum of n terms of the series (4/3)+(10/9)+(28/27)+... is:

Question

The sum of n terms of the series (4/3)+(10/9)+(28/27)+... is: (A)  (3n(2n+1)+1/2(3n))  (B)  (3n(2n+1)-1/2(3n))  (C)  (3nn-1/2(3n))  (D)  (3n-1/2)

Tardigrade Admin · Accepted Answer

The given series can be rewritten as ( 1+(1/3) )+( 1+(1/9) )+( 1+(1/27) )+...n =n+(1/3)( 1+(1/3)+(1/32)+...n textterms ) =n+(1/3)([ 1-( (1/3) )n ]/1-(1)3) =n+((1/3)( 1-(1/3n) )/(2)3) =(2n+1-(1/3n)/2)=(3n(2n+1)-1/2⋅ 3n)

Q. The sum of n terms of the series $\frac{4}{3} + \frac{10}{9} + \frac{28}{27} + ...$ is:

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

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

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

$\frac{3 ^{n} - 1}{2}$

Q. The sum of n terms of the series 34​+910​+2728​+... is:

2(3n)3n(2n+1)+1​

2(3n)3n(2n+1)−1​

2(3n)3nn−1​

23n−1​

The given series can be rewritten as (1+31​)+(1+91​)+(1+271​)+...n =n+31​(1+31​+321​+...nterms) =n+31​1−31​[1−(31​)n]​ =n+32​31​(1−3n1​)​ =22n+1−3n1​​=2⋅3n3n(2n+1)−1​

Q. The sum of n terms of the series $\frac{4}{3} + \frac{10}{9} + \frac{28}{27} + ...$ is:

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

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

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

$\frac{3 ^{n} - 1}{2}$