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Mathematics Question on Geometric Progression

The sum of the first 10 terms of the series 9 + 99 + 999 + ?., is

A

98(910−1)\frac{9}{8} \left(9^{10} - 1\right)

B

1009(109−1)\frac{100}{9} \left(10^{9} - 1\right)

C

109−110^9 - 1

D

1009(1010−1)\frac{100}{9} \left(10^{10} - 1\right)

Answer

1009(109−1)\frac{100}{9} \left(10^{9} - 1\right)

Explanation

Solution

Let, Sn=9+99+999+……n S_{n}=9+99+999+\ldots \ldots n terms
⇒Sn=(10−1)+(100−1)+(1000−1)+…n\Rightarrow S_{n}=(10-1)+(100-1)+(1000-1) +\ldots n terms
⇒Sn=(10+102+103+……n\Rightarrow S_{n}=\left(10+10^{2}+10^{3}+\ldots \ldots n\right. terms ))
−(1+1+……n-( 1 + 1 + \dots \dots n terms ))
⇒Sn=10(10n−1)10−1−n\Rightarrow S_{n}=\frac{10\left(10^{n}-1\right)}{10-1}-n
[∵a+ar+ar2+……+arn−1=a(rn−1)r−1,r>1]\left[\because a+a r+a r^{2}+\ldots \ldots+a r^{n-1}=\frac{a\left(r^{n}-1\right)}{r-1}, r>1\right]
⇒Sn=109(10n−1)−n\Rightarrow S_{n}=\frac{10}{9}\left(10^{n}-1\right)-n
Put n=10n=10
⇒S10=109(1010−1)−10\Rightarrow S_{10} =\frac{10}{9}\left(10^{10}-1\right)-10
=109(1010−1−9)=\frac{10}{9}\left(10^{10}-1-9\right)
=109(1010−10)=1009(109−1)=\frac{10}{9}\left(10^{10}-10\right)=\frac{100}{9}\left(10^{9}-1\right)