If the sum of the series 1+3/2+5/4+7/8+....+(2n-1)/(2^(n-1)) is fn, then the value of f8 is

Correct answer: (C) 6-19/(2^7). S_n=1+3/2+5/4+7/8+.....+(2n-1)/(2^(n-1)) 1/2S_n=1/2+3/4+5/8+.....+(2n-3)/(2^(n-1))+(2n-1)/(2^n) Subtracting, we get, 1/2S_n=1+2/2+2/4+2/8+.....+2/(2^(n-1))-(2n-1)/(2^n) 1/2S_n=1+21/2+1/4+1/8+.....n-1terms-(2n-1)/(2^n) 1/2S_n=1+21/21-1^(2n-1)1-1/2-(2n-1)/(2^n) 1/2S_n=1+21-1/(2^(n-1))-(2n-1)/(2^n) =1+2-1/(2^(n-2))-(2n-1)/(2^n) fn=S_n=6-1/(2^(n-3))-(2n-1)/(2^(n-1))⇒f8=6-1/(2^5)-15/(2^7) f8=6-(4+15)/(2^7)=6-19/(2^7)

Q49 / 118 · NTA Abhyas (JEE Main) 2020 · Mathematics · Sequences and Series

If the sum of the series $1 + \frac{3}{2} + \frac{5}{4} + \frac{7}{8} + . . . . + \frac{(2 n - 1)}{{(2)}^{n - 1}}$ is $f (n),$ then the value of $f (8)$ is

A
$4 + \frac{12}{2^{5}}$
B
$5 + \frac{13}{2^{7}}$
Correct answer
C
$6 - \frac{19}{2^{7}}$
D
$5 - \frac{13}{2^{7}}$

Solution

S_{n} = 1 + \frac{3}{2} + \frac{5}{4} + \frac{7}{8} + . . . . . + \frac{(2 n - 1)}{2^{n - 1}}

\frac{1}{2} S_{n} = \frac{1}{2} + \frac{3}{4} + \frac{5}{8} + . . . . . + \frac{(2 n - 3)}{2^{n - 1}} + \frac{(2 n - 1)}{2^{n}}

Subtracting, we get,

\frac{1}{2} S_{n} = 1 + \frac{2}{2} + \frac{2}{4} + \frac{2}{8} + . . . . . + \frac{2}{2^{n - 1}} - \frac{(2 n - 1)}{2^{n}}

\frac{1}{2} S_{n} = 1 + 2 (\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . . . (n - 1) t e r m s) - \frac{(2 n - 1)}{2^{n}}

\frac{1}{2} S_{n} = 1 + 2 \frac{\frac{1}{2} (1 - {(\frac{1}{2})}^{n - 1})}{(1 - \frac{1}{2})} - \frac{(2 n - 1)}{2^{n}}

\frac{1}{2} S_{n} = 1 + 2 (1 - \frac{1}{2^{n - 1}}) - \frac{(2 n - 1)}{2^{n}}

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

f (n) = S_{n} = 6 - \frac{1}{2^{n - 3}} - \frac{(2 n - 1)}{2^{n - 1}} \Rightarrow f (8) = 6 - \frac{1}{2^{5}} - \frac{15}{2^{7}}

f (8) = 6 - \frac{(4 + 15)}{2^{7}} = 6 - \frac{19}{2^{7}}

Hence, the correct answer is option: 'C'.

If the sum of the series 1+32+54+78+....+2n-12n-1 is fn, then the value of f8 is

Solution

More Sequences and Series Questions

If the sum of the series $1 + \frac{3}{2} + \frac{5}{4} + \frac{7}{8} + . . . . + \frac{(2 n - 1)}{{(2)}^{n - 1}}$ is $f (n),$ then the value of $f (8)$ is