8. Sequences and Series
hard

નીચેની શ્રેણીનાં પ્રથમ $n$ પદોનો સરવાળો શોધો :

$6+.66+.666+\ldots$

A

$\frac{2}{3} n-\frac{2}{27}\left(1-10^{-n}\right)$

B

$\frac{2}{3} n-\frac{2}{27}\left(1-10^{-n}\right)$

C

$\frac{2}{3} n-\frac{2}{27}\left(1-10^{-n}\right)$

D

$\frac{2}{3} n-\frac{2}{27}\left(1-10^{-n}\right)$

Solution

$6+.66+.666+\ldots$

Let $S_{n}=06+0.66+0.666+\ldots .$ to $n$ terms

$=6[0.1+0.11+0.111+\ldots . \text { to } n \text { terms }]$

$=\frac{6}{9}[0.9+0.99+0.999+\ldots . . \text { to } n \text { terms }]$

$=\frac{6}{9}\left[\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{10^{2}}\right)+\left(1-\frac{1}{10^{3}}\right)+\ldots . \text { to } n \text { terms }\right]$

$=\frac{2}{3}\left[(1+1+\ldots n \text { terms })-\frac{1}{10}\left(1+\frac{1}{10}+\frac{1}{10^{2}}+\ldots n \text { terms }\right)\right]$

$=\frac{2}{3}\left[n-\frac{1}{10}\left(\frac{1-\left(\frac{1}{10}\right)^{n}}{1-\frac{1}{10}}\right)\right]$

$=\frac{2}{3} n-\frac{2}{30} \times \frac{10}{9}\left(1-10^{-n}\right)$

$=\frac{2}{3} n-\frac{2}{27}\left(1-10^{-n}\right)$

Standard 11
Mathematics

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