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8. Sequences and Series
normal
જો ${s_n} = 1 + \frac{1}{2} + \frac{1}{{{2^2}}} + ........ + \frac{1}{{{2^{n - 1}}}}$ ,હોય તો $n$ ની ન્યૂનતમ કિમત મેળવો કે જેથી $2 - {s_n} < \frac{1}{{100}}$ થાય
A
$7$
B
$9$
C
$8$
D
$6$
Solution
$S_{n}=1+1 / 2+1 / 2^{2}+\ldots \ldots \ldots+1 / 2^{n-1}$
$S_{n}=\frac{1\left(1-(1 / 2)^{n}\right)}{(1-1 / 2)}=2\left[1-\frac{1}{2^{n}}\right]$
$2-\mathrm{Sn}<\frac{1}{100}$
$\frac{2}{2^{n}}<\frac{1}{100}$
$n \geq 8$
Standard 11
Mathematics
