The sum of $n$ terms of the series $5+7+13+31+85+\ldots$ is

$\frac{1}{2}\left(3^n+7 n+1\right)$

$\frac{1}{2}\left(3^n+5 n\right)$

$\frac{1}{2}\left(3^n+7 n\right)$

Back to SEQUENCE OF SERIES

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Medium

1 of 1

Application +4 | -1

The sum of

n

terms of the series

5+7+13+31+85+\ldots

View answer and explanation

Correct option: A

\frac{1}{2}\left(3^n+8 n-1\right)

Option A (

\frac{1}{2}(3^n+8n-1)

) is correct because the sum of n terms of the series is derived to be in this form by identifying the pattern in the differences of successive terms.

More Questions from SEQUENCE OF SERIES

Practice All Chapter MCQs →

Given the sets

A=\{1,2,3\}, B=\{3,4\}, C=\{4,5,6\}

, then find the following
(i)

{A} \cup({B} \cap {C})

(ii)

A-(B \cap C)

(iii)

(B \cup C)-A

a,b,c

are in A.P.;

a,x,b

are in G.P. and

b,y,c

are in G.P., then

{x^2},{b^2},{y^2}

are in Suppose

P(x)=1-x+x^2-x^3+\ldots+x^{2012}

is expressed as a polynomial in

y

, as

Q(y)=a_0+a_1 y+\ldots+a_{2012} y^{2012}

where

y=x-2

, then

\sum\limits_{{i=0}}^{2012} a_i

equals If

\log \left(\frac{5 c}{a}\right), \log \left(\frac{3 b}{5 c}\right)

and

\log \left(\frac{a}{3 b}\right)

are in

A P

, where

a, b

and c are in

G P

, then

a, b, c

are the lengths of sides of If the sum of three numbers in G.P. is 63 and the product of the first and the second term is

\frac{3}{4}

of the third term, then the numbers are If the lengths of sides of a right angled triangle are in A.P., then their ratio is